IAT SOLUTIONS - C_7.pdf
IAT SOLUTIONS - C_7.pdf
IAT SOLUTIONS - C_7.pdf
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Chapter 7Copyright © by McDougal Littell, a division of Houghton Mifflin Company.Prerequisite Skills (p. 476)1. The domain of the function is x ≥ 2.2. The range of the function is y ≥ 3.3. The inverse of the function is y 5 (x 2 3) 2 1 2.4. y 5 22 Ï } x 2 1 5. y 5 Ï } x 1 3Domain: x ≥ 0 Domain: x ≥ 23Range: y ≤ 21 Range: y ≥ 021y16. y 5 3 Ï } x 2 2 1 5 7. y 5 3x 1 5xDomain: all real numbers x 5 3y 1 5Range: all Real numbers211yx211yx 2 5 5 3yx 2 5} 5 y38. y 5 22x 3 1 1 9. y 5 1 }2 x2 , x ≥ 0x 5 22y 3 1 1 x 5 }1 2 y2x 2 1 5 22y 3 2x 5 y 22 }x 2 1 5 y23 Ï } 2x 5 y3Î } 2 }x 2 1 5 y210. y 5 ax 2 1 bx 1 c(0, 21): 21 5 a(0) 2 1 b(0) 1 c → c 5 21(1, 2): 2 5 a(1) 2 1 b(1) 1 c → a 1 b 1 c 5 2(3, 14): 14 5 a(3) 2 1 b(3) 1 c → 9a 1 3b 1 c 5 14a 1 b 1 (21) 5 2 → a 1 b 5 3 → a 5 3 2 b9a 1 3b 1 (21) 5 14 → 9a 1 3b 5 159(3 2 b) 1 3b 5 15 → b 5 2a 5 3 2 2 5 1An equation for the parabola is y 5 x 2 1 2x 2 1.11. y 5 ax 2 1 bx 1 c(3, 8): 8 5 a(3) 2 1 b(3) 1 c → 9a 1 3b 1 c 5 8(4, 17): 17 5 a(4) 2 1 b(4) 1 c → 16a 1 4b 1 c 5 17(7, 56): 56 5 a(7) 2 1 b(7) 1 c → 49a 1 7b 1 c 5 569a 1 3b 1 c 5 8 3 (21) 29a 2 3b 2 c 5 2816a 1 4b 1 c 5 17 16a 1 4b 1 c 5 177a 1 b 5 9x16a 1 4b 1 c 5 17 3 (21) 216a 2 4b 2 c 5 21749a 1 7b 1 c 5 56 49a 1 7b 1 c 5 5633a 1 3b 5 397a 1 b 5 9 3 (23) 221a 2 3b 5 22733a 1 3b 5 39 33a 1 3b 5 397(1) 1 b 5 9 → b 5 29(1) 1 3(2) 1 c 5 8 → c 5 2712a 5 12a 5 1An equation for the parabola is y 5 x 2 1 2x 2 7.12. y 5 ax 2 1 bx 1 c(23, 9): 9 5 a(23) 2 1 b(23) 1 c → 9a 2 3b 1 c 5 9(1, 27): 27 5 a(1) 2 1 b(1) 1 c → a 1 b 1 c 5 27(5, 255): 255 5 a(5) 2 1 b(5) 1 c → 25a 1 5b 1 c5 255a 1 b 1 c 5 27 → c 5 2a 2 b 2 79a 2 3b 1 (2a 2 b 2 7) 5 9 → 8a 2 4b 5 1625a 1 5b 1 (2a 2 b 2 7) 5 255 → 24a 1 4b 5 2488(21) 2 4b 5 16 → b 5 26(21) 1 (26) 1 c 5 27 → c 5 0An equation for the parabola is y 5 2x 2 2 6x.Lesson 7.17.1 Guided Practice (pp. 479–481)32a 5 232a 5 211. Domain: all real numbers 2. Domain: all real numbersRange: y > 0 Range: y > 0212y3. Domain: all real numbersRange: y > 2(21, 3)f (x)f(x) 5 3 x 1 1 1 2(0, 5)f(x) 5 3 x 221(0, 1)(1, 3)xx4. Using the graph, you can estimate that the number ofincidents was about 250,000 during 2003 (t ø 7).5. In the exponential growth model y 5 527(1.39) x , theinitial amount is 527, the growth factor is 1.39, and thepercent increase is 1 2 1.39 5 0.39 or 39%.211yxAlgebra 2Worked-Out Solution Key419
Chapter 7, continued6. When p 5 2000, r 5 0.04, t 5 3, and n 5 365:A 5 P 11 1 } n r 2 ntA 5 2000 1 1 1 0.04 }365 2 (365)(3)5 20001 }365.04365 2 1095ø 2254.98The balance after 3 years is $2254.98.7.1 Exercises (pp. 482–485)Skill Practice1. In the exponential growth model y 5 2.4(1.5) x , the initialamount is 2.4, the growth factor is 1.5, and the percentincrease is 1 2 1.5 5 0.5 or 50%.2. An asymptote is a line that a graph approaches more andmore closely.3. C; When x 5 0, y 5 3 p 2 0 5 3;When x 5 1, y 5 3 p 2 1 5 64. A; When x 5 0, y 5 23 p 2 0 5 23;When x 5 1, y 5 23 p 2 1 5 265. B; When x 5 0, y 5 2 p 3 0 5 2;When x 5 1, y 5 2 p 3 1 5 66.y7.y212x11xy 5 5 ? 4 x14.h(x)11xy15.16.y1y 5 23 ? 2 x (0, 7)1 x13 (0, 5)(0, 23)( 21,4 )(22, 23)(1, 26) y 5 5 ? 4 x 1 2y 5 23 ? 2 x 1 21(21, 26)5 1 x( 21,4)Domain: all real number Domain: all real numbersRange y < 0 Range: y > 217.y18.y19.y 5 2 x 1 1 1 3(1, 3)(21, 4)2(3, 2)y 5 3 x (0, 1)722,2 y 5 2 x221 (2, 0) x(0, 1)y 5 3 x 2 2 2 11 1 x21,2( )( )Domain: all real numbers Domain: all real numbersRange: y > 3 Range: y > 21y(1, 6)(3, 5)Domain: all real numbersRange: y > 218.10.12.f(x) 9.121xy221xy11.13.y221g(x)11yxx20.21.3(0, 2)y 5 2 ? 3 x (2, 1)21xy 5 2 ? 3 x 2 2 2 1yy 5 23 ? 4 x 1(21, 20.75)1 x(0, 22.75)Domain: all real numbersRange: y < 22f(x) Domain: all real numbersRange: y > 3(21, 3) 2f(x) 5 6 ? 2 x 2 x(0, 23)(1, 25)y 5 23 ? 4 x 2 1 2 2f(x) 5 6 ? 2 x 2 3 1 3(3, 9)(0, 6)(2, 6)Copyright © by McDougal Littell, a division of Houghton Mifflin Company.1121x21x420Algebra 2Worked-Out Solution Key
Chapter 7, continued37. A 5 2200 1 1 1 r } n 2 n(4) ; P 5 2200; t 5 4a. r 5 0.03, n 5 4A 5 2200 1 1 1 0.03 }4 2 4 p 4 5 2200(1.0075) 16 ø 2479.38The balance after 4 years is $2479.38.b. r 5 0.0225, n 5 12A 5 2200 1 1 1 0.0225 }12 2 12 p 4 5 2200(1.001875) 48ø 2406.98The balance after 4 years is $2406.98.c. r 5 0.02, n 5 365A 5 2200 1 1 1 0.02 }365 2 365 p 4 5 2200(1.00005479452) 1460ø 2383.23The balance after 4 years is $2383.23.38. a. When A 5 3000, r 5 0.0225, n 5 4, and t 5 3:A 5 p 11 1 }n2 r nt3000 5 p 11 1 }0.02254 2 4 p 33000 5 p (1.005625) 122804.71 ø pb. When A 5 3000, r 5 0.035, n 5 12, and t 5 3:A 5 p 11 1 } n r 2 nt3000 5 p 11 1 }0.03512 2 12 p 33000 5 p 1 }12.03512 2 362701.39 ø pYou should deposit $2701.39 to have $3000 in youraccount after 3 years.c. When A 5 3000, r 5 0.04, n 5 1, and t 5 3:A 5 p 11 1 }n2 r nt3000 5 p 11 1 }0.041 2 1 p 33000 5 p (1.04) 32666.99 ø pYou should deposit $2666.99 to have $3000 in youraccount after 3 years.39. a. Initial amount 5 494.29 thousand; percent increase5 0.03; growth factor 5 1 1 0.03 5 1.03Exponential growth model: P 5 494.29(1.03) tWhen t 5 10: P 5 494.29(1.03) 10 ø 664.284The population in 2000 was about 664,284 people.b. Domain: 0 ≤ t ≤ 10Range: 494.29 ≤ P ≤ 664,284Population (thousands)P70060050040030020010000 1 2 3 4 5 6 7 8 9 tYear (t 5 0; 1990)c. Using the graph, you can estimate that the populationwas about 590,000 during 1996 (t ø 6).40. a. Initial amount 5 50; percent increase 5 0.105;number of bids 5 nExponential growth model: p 5 50(1 1 0.105) n→ p 5 50(1.105) nb. When n 5 5: p 5 50(1.105) 5 ø 82.37. The priceafter 5 bids was $82.37. After 100 bids (n 5 100),the price will be p 5 50(1.105) 100 ø $1,084,420.72.No. Sample answer: This amount is unreasonablebecause the model is only defined for 6 bids and100 is out of this domain.41. a. Initial amount 5 48.28; percent increase 5 0.06Exponential growth model: p 5 48.28(1 1 0.06) t→ p 5 48.28(1.06) tb. Using the graph, you can estimate that the price of aticket was $60 during 2004 (t ø 4).Average price (dollars)p807060504030201000 1 2 3 4Years (t 5 0; 2000)c. Over the domain t ≥ 0, the minimum value of t ist 5 0. The function is defined in the interval0 ≤ t ≤ 4. You can look at the graph between thesevalues to determine the minumum or maximum thatgives meaningful results.42. a. Initial amount 5 41; percent increase 5 0.089Exponential growth model: n 5 41(1 1 0.089) t→ n 5 41(1.089) ttCopyright © by McDougal Littell, a division of Houghton Mifflin Company.422Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.b. t nc.Number of breeding pairs0 411 44.6492 48.6233 52.9504 57.6635 62.7956 68.3837 74.4708 81.097n330300270240210180150120906030tYears since 1977n9 88.31510 96.17511 104.73512 114.05613 124.20714 135.26215 147.23016 160.41000 3 6 9 12 15 18 21 24ttn17 174.68618 190.23319 207.16420 225.60121 245.68022 267.54523 291.35724 317.288d. Using the graph, you can estimate that in 2001 (t 5 24),there were about 315 breeding pairs of bald eagles.43. Investing $3000 at 6% annual interest: A 15 3000(1.06) tInvesting $3000 at 8% annual interest: A 25 3000(1.08) tInvesting $6000 at 7% annual interest: A 35 6000(1.07) tx 1 2 3 4A 11 A 26420 6870 7352.18 7868.90A 36420 6869.40 7350.26 7864.78No, A 11 A 2and A 3are not the same. After the first year,the money split between the 6% and 8% accounts growsat a faster rate.44. a. Initial amount 5 5200; final amount 5 9000; numberof years 5 59000 5 5200(1 1 r) 545}26 5 (1 1 r)5 → Î } 45 5 }26 5 1 1 r0.116 ø rThe average annual growth rate was about 11.6%b. y 5 9000(1 1 0.116) 5 5 9000(1.116) 5 ø 15,579.86The cost will be $15,579.86 in 5 more years.47. (0.5) 5 5 (0.5)(0.5)(0.5)(0.5)(0.5) 5 0.0312548. (0.25) 3 5 (0.25)(0.25)(0.25) 5 0.01562549. 1 }1 22 4 5 1 }1 221 }1 221 }1 221 }1 22 5 }1 1650. 1 }3 82 3 5 1 }3 821 }3 821 }3 82 5 }27 51251. 1 }102 7 5 5 1 }102 7 1 }102 7 1 }102 7 1 }102 7 1 }752. 1 }4 52 3 5 1 }4 521 }4 521 }4 52 5 }64 125102 5 }16,807100,00053. x 2 1 7x 2 30 5 (x 1 10)(x 2 3)54. x 2 1 15x 1 54 5 (x 1 9)(x 1 6)55. 2x 2 2 7x 2 30 5 (2x 1 5)(x 2 6)56. 12x 2 2 5x 1 25 cannot be factored57. x 3 2 2x 2 2 3x 1 6 5 x 2 (x 2 2) 2 3(x 2 2)5 (x 2 2 3)(x 2 2)58. x 3 2 64 5 (x 2 4)(x 2 1 4x 1 16)59. x 5 5 3125 60. 3x 3 5 1029x 5 5 x 3 5 343x 5 761. x 7 1 8 5 264 62. (x 1 12) 4 5 52x 7 5 272 (x 1 12) 5 6 Ï 4 } 52x 5 Ï 7 } 272 x 5 2126 Ï 4 } 5263. 25x 6 5 21000 64. (x 2 9) 8 5 17x 6 5 200 x 2 9 5 6 Ï 8 } 17x 5 6 Ï 6 } 200 x 5 96 Ï 8 } 17Lesson 7.27.2 Guided Practice (pp. 487–488)1.y2.3.1f(x)21221 xx4.(21, 4)21(0, 1)21y1y(0, 5)y 51(1, 2)x1( ) x 2 1 4 1 1y 51( 4)Domain: all real numbersRange: y > 1xxMixed Review45. (0.6) 3 5 (0.6)(0.6)(0.6) 5 0.21646. (0.4) 2 5 (0.4)(0.4) 5 0.16Algebra 2Worked-Out Solution Key423
Chapter 7, continued5.y(0, 5)10( 1,3 )2 x(21, 3) y 5 5 ( 3)4( 0,3 )211y 5 5x2( ) x 1 1 3 2 26.g(x)21g(x) 5 2323( ) x 2 5 4 1 491, 24(0, 23)3 xg(x) 5 23 4( )76,4 x(5, 1) ( )( )7.9.211yf(x)x8.10.y121 xy121xDomain: all real numbers Domain: all real numbersRange: y > 22 Range: y < 47. Initial amount 5 4200; percent decrease 5 0.20Exponential decay model:y 5 4200(1 2 0.20) t → y 5 4200(0.80) t using the graph,you can estimate that the value of the snowmobile will be$2500 after about 2 years.y400011.21121y1xx12.211g(x)xValue (dollars)30002000100000 1 2 3 4 5 6 7 8 9 tTime (yr)8. Final cost 5 3000; number of years 5 3; percentdecrease 5 0.073000 5 a(1 2 0.07) 33000 5 a(0.93) 33729.69 ø aThe original cost of the snowmobile was $3729.69.7.2 Exercises (pp. 489–491)Skill Practice1. Initial amount 5 1250; decay factor 5 0.85; percentdecrease 5 1 2 0.85 5 0.15 or 15%2. The function y 5 b x represents exponential growth ifb > 1, and exponential decay if 0 < b < 1.3. f (x) 5 31 }3 42 x , b 5 }3 4Because 0 < b < 1, thisfunction represents exponential decay.4. f (x) 5 41 }5 22 x , b 5 }5 2represents exponential growth.Because b > 1, this function5. f (x) 5 }2 7 p 4x , b 5 4 Because b > 1, this functionrepresents exponential growth.6. f (x) 5 25(0.25) x , b 5 0.25 Because 0 < b < 1,this function represents exponential decay.13.2115. B; y 5 221 }3 52 x16.2yx14.21h(x)1When x 5 0: y 5 221 }3 52 0 5 22(1) 5 22 (0, 22)When x 5 1: y 5 221 }3 52 1 5 221 }3 52 5 2 }6 5 11, 2 }6 52(0, 1)2121y(1 xy 5 3 1 1(0, 2)1( 1,3 )(4( 1,3 )(1 xy 5 3(x17.(y11( 1, 22 )24 (0, 21)1 xy 5 2 ( 2(1, 21)1( 2, 22 )(x1 x 2 1y 5 2 2Domain: all real numbers Domain: all real numbersRange: y > 1 Range: y < 018.y19.y2(0, 2) 1,31 x1y 5 2 ( 3(21, 21) 5 x1 x 1 1y 5 2 ( 3 2 370, 23( )( )(((y 5(2 x32( 1,3 )(2(0, 1)y 51((2 x 2 43 2 1(4, 0)1( 5, 23 )Domain: all real numbers Domain: all real numbersRange: y > 23 Range: y > 21xxCopyright © by McDougal Littell, a division of Houghton Mifflin Company.424Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.20.(0, 6)(0, 3)1yy 5 3(0.25) x 1 315( 1,4 )3( 1,4 )1y 5 3(0.25) x x21.y3(0, 1)21(1 x 2 2y 5 3 1 21( 1,3 )(2, 3)(7(3,3 )(1 x xy 5 3Domain: all real numbers Domain: all real numbersRange: y > 3 Range: y > 222.f(x)x 23.g(x)1( )31, 24(0, 23)1f(x) 5 23 453( 2, 24 )(1, 23)f(x) 5 231( 4)( )x 2 1x(24, 1)g(x) 5 6g(x) 5 6(25, 4)1( 2)1( 2)x1x 1 5 2 2(0, 6)1(1, 3)Domain: all real numbers Domain: all real numbersRange: y < 0 Range: y > 2224.h(x)(21, 4)(0, 2)121h(x) 5 4(0, 4)(1, 2)1( 2)( )x1 x 1 1 xh(x) 5 4 2Domain: all real numbersRange: y > 025. y 5 ab x 2 h 1 k → y 5 3(0.4) x 2 2 2 1a. If a changes to 4, there will be a vertical stretch.b. If b changes to 0.2, the graph will be steeper becausethe decay factor is smaller.c. If h changes to 5, there will be a horizontal translation(of 3 units right)d. If k changes to 3, there will be a vertical translation(of 4 units up).26. The decay factor was written incorrectly. It should be1 2 percent decrease 5 1 2 0.02 5 0.98.y 5 (inital amount)(decay factor) ty 5 500(0.98) t27. D; The graph y 5 1 }1 22 x 2 2 1 3 has an asymptote at theline y 5 3.28. You need a base between 0.25 and 0.5 and a verticaltranslation up between 0 and 3 units. Sample answer:y 5 (0.3) x 1 129. Yes; (4) 2x is just another way of saying 1 }1 42 x or(0.25) x , so f (x) 5 5(4) 2x and g(x) 5 5(0.25) x representthe same function.(xProblem Solving30. a. When I 5 200 and t 5 1.5: A 5 I(0.71) tA 5 200(0.71) 15 ø 119.65There is about 120 milligrams of ibuprofen remainingin the bloodstream.b. When I 5 325 and t 5 3.5: A 5 I(0.71) tA 5 325(0.71) 3.5 ø 98.01There is about 98 milligrams of ibuprofen remainingin the bloodstream.c. When I 5 400 and t 5 5: A 5 I(0.71) tA 5 400(0.71) 5 ø 72.17There is about 72 milligrams of ibuprofen remainingin the bloodstream.31. a. When r 5 0.25 and a 5 200: y 5 a(1 2 r) ty 5 200(1 2 0.25) t → y 5 200(0.75) tWhen t 5 3: y 5 200(0.75) 3 ø 84.38The value of the bike after 3 years is about $84.38.c. Using the graph, you can estimate that the value of thebike will be $100 after about 2.5 years.Value (dollars)y20017515012510075502500 1 2 3 4 5 6 7 8 tYears32. Ratio from year 1–2: 1832 }1906 ø 0.96Ratio from year 2–3: 1762 }1832 ø 0.96Ratio from year 3–4: 1692 }1762 ø 0.96Ratio from year 4–5: }16271692 ø 0.96The ratio of depreciation remains a constant 0.96 or 96%.d 5 a(0.96) 1 5 1906 → a 5 }19060.96 ø 1985An equation is d 5 1985(0.96) t .33. a. yValue (dollars)28,00024,00020,00016,00012,0008000400000 1 2 3 4 5 6 7 8 tYearsUsing the graph, you can estimate that the value of thecar will be $10,000 after about 5 years.Algebra 2Worked-Out Solution Key425
Chapter 7, continuedb. When t 5 50: y 5 24,000(0.845) 50 ø 5.29The value after 50 years is $5.29 according tothe model. Sample answer: This is too low to bereasonable for the price of a car. In addition, car valuesusually begin to increase once they become antiques.34. a. P 5 1001 }1 22 t /5730When t 5 2500: P 5 1001 }1 22 2500/5730 ø 73.90When t 5 5000: P 5 1001 }1 22 5000/5730 ø 54.62When t 5 10,000: P 5 1001 }1 22 10,000/5730 ø 29.83After 2500 years there will be about 73.9% of theoriginal carbon-14 remaining, after 5000 years therewill be about 54.6% remaining, and after 10,000 yearsthere will be about 29.8% remaining.b. PPercent100908070605040302010004000800012,00016,000Yearsc. Using the graph, you can estimate the age of the bisonbone is about 8000 years old when 37% of thecarbon-14 is present.35. a. The decay facter is 0.89 and the percent decrease is1 2 0.89 5 0.11, or 11%.b. ENumber of eggsproduced per year18017016015014013012011010000 20 40 60 80 100 120 140wAge of chicken (weeks)c. There are 52 weeks in a year, so when a chicken is2.5 years old, it is 2.5(52) 5 130 weeks old. Usingthe graph, you can estimate that the number of eggsproduced by a 130 week old chicken is about 134per year.d. Let t represent the chicken’s age in years.Because t 5 }w , you can rewrite the equation as52E 5 179.2(0.89) t .t36. Initial amount 5 1300; resale value after t years 5 275;number of years 5 4V 5 a(1 2 r) t275 5 1300(1 2 r) 40 p 2115 ø (1 2 r) 40.678 ø 1 2 rThe decay factor is 0.678, so an equation giving thestereo’s resale value as a function of time isV 5 1300(0.678) t .Mixed Review37.38.39.40.41.42.21232321122y1y11yyy21y21 xxxxxxDomain and range:all real numbersDomain: all real numbersRange: y ≥ 22.25Domain: all real numbersRange: y ≥ 29Domain: x ≥ 0Range: y ≥ 3Domain: x ≥ 4Range: y ≥ 0Domain and range:all real numbersCopyright © by McDougal Littell, a division of Houghton Mifflin Company.426Algebra 2Worked-Out Solution Key
Chapter 7, continued43.44.45.y121 xy123 xDomain: all real numbersRange: y > 0Domain: all real numbersRange: y < 0y Domain: all real numbersRange: y > 0Lesson 7.37.3 Guided Practice (pp. 493–495)1. e 7 p e 4 5 e 7 1 4 5 e 112. 2e 23 p 6e 5 5 12e 23 1 5 5 12e 23. 24e8 }4e 5 5 6e8 2 5 5 6e 34. (10e 24x ) 3 5 10 3 (e 24x ) 3 5 1000e 212x 5 1000 }e 12x5. e 3/4 ø 2.1176.1yDomain: all real numbersRange: y > 0Copyright © by McDougal Littell, a division of Houghton Mifflin Company.121 x46. f (x) 5 5x 2 2, g(x) 5 x 1 2 }5f ( g (x)) 5 5 1 x 1 2 }5 2 2 2 5 x 1 2 2 2 5 x(5x 2 2) 1 2g ( f (x)) 5 }55 5x }5 5 xBecause f ( g(x)) 5 x and g ( f (x)) 5 x, f and g are inversefunctions.47. f (x) 5 23x 1 10, g (x) 5 10 2 x }3f ( g (x)) 5 23 1 10 2 x }3 2 1 10 5 210 1 x 1 10 5 x10 2 (23x 1 10) 10 1 3x 2 10g ( f (x)) 5 }} 5 }}335 3x }3 5 xBecause f ( g (x)) 5 x and g ( f (x)) 5 x, f and g are inversefunctions.48. f (x) 5 4x 3 2 7, g(x) 5 1 }x 1 74 2 1/3f ( g (x)) 5 4 1 1 }x 1 74 2 1/3 2 3 2 7 5 41 }x 1 75 x 1 7 2 7 5 x4 2 2 7g ( f (x)) 5 1 (4x3 2 7) 1 7 1/3}}42 5 1 }4x342 1/3 5 (x 3 ) 1/3 5 xBecause f ( g (x)) 5 x and g ( f (x)) 5 x, f and g are inversefunctions.49. f (x) 5 x5 1 7}, g (x) 5 5 12Ï } 12x 2 7( Ï 5} 12x 2 7 ) 5 1 7f ( g (x)) 5 }} 5 }12x 2 7 1 7 5 }12x1212 12 5 x5Î}}g ( f (x)) 5 121 x5 1 7}12 2 2 7 5 Ï 5 } x 5 1 7 2 7 5 Ï 5 } x 5 5 xBecause f ( g (x)) 5 x and g ( f (x)) 5 x, f and g are inversefunctions.7.8.9.1( 0,2 )21322(1, 0.18)f(x)1( 2 )f(x) 5 e 2x 1 13( 0,2 )(1, 1.18)( )1 xf(x) 5 e 2x2(1, 1.93)3( 0,2 )y 5 1.5e 0.25x 1211( 1, 22 )y 5 1.5e 0.25(x 2 1) 2 2X=5 Y=223.65901yxx(2, 20.07)Domain: all real numbersRange: y > 1Domain: all real numbersRange: y > 22Use the trace feature to determine that l ø 224when t 5 5.The length of a 5-year-old tiger shark is about224 centimeters.10. A 5 Pe rt where p 5 2500 and r 5 0.05a. When t 5 2: A 5 2500e 0.05(2) 5 2500e 0.1 ø 2762.93The balance after 2 years is $2762.93.b. When t 5 5: A 5 2500e 0.05(5) 5 2500e (0.25) ø 3210.06The balance after 5 years is $3210.06c. When t 5 7.5: A 5 2500e 0.05(7.5)5 2500e 0.375 ø 3637.48The balance after 7.5 years is $3637.48.Algebra 2Worked-Out Solution Key427
Chapter 7, continued11. Amount of interest earned 5 balance 2 principle19. e 3 ø 20.086 20. e 23/4 ø 0.472e}6xe 5 22x e6x 2 (22x) 5 e 8xa. The interest earned after 2 years is21. e 2.2 ø 9.025 22. e 1/2 ø 1.6492762.93 2 2500 5 $262.93.23. e 22/5 ø 0.670 24. e 4.3 ø 73.700b. The interest earned after 5 years is25. e 7 ø 1096.633 26. e 24 ø 0.0183210.06 2 2500 5 $710.06.27. 2e 20.3 ø 1.482 28. 5e 2/3 ø 9.739c. The interest earned after 7.5 years is29. 26e3637.48 2 2500 5 $1137.48.ø 266.139 30. 0.4e 4.1 ø 24.13631. The function f (x) 5 3e 2x is an example of exponential7.3 Exercises (pp. 495–498)Skill Practicedecay.32. The function f (x) 5 }1 3 e4x is an example of exponentialgrowth.1. The number e is an irrational number approximately33. The function f (x) 5 e 24x is an example of exponentialequal to 2.71828.2. The function f (x) 5 }1 decay.3 e4x is an example of exponential 34. The function f (x) 5 }3growth because }1 5 ex is an example of exponentialgrowth.> 0 and 4 > 0.335. The function f (x) 5 }13. e 3 p e 4 5 e 3 1 4 5 e 74 e25x is an example of exponentialdecay.4. e 22 p e 6 5 e 22 1 6 5 e 45. (2e 3x ) 3 5 2 3 (e 3x ) 3 5 8e 9x36. The function f (x) 5 e 3x is an example of exponentialgrowth.6. (2e 22 ) 24 5 2 24 (e 22 ) 24 5 }1 16 e8 5 }e837. The function f (x) 5 2e 4x is an example of exponential16growth.7. (3e 5x ) 21 5 3 21 (e 5x ) 21 5 }1 138. The function f (x) 5 4e 22x is an example of exponential3 e25x 5 }3 e 5xdecay.8. e x p e 23x p e 4 5 e x 2 3x 1 4 5 e 22x 1 49. Ï } 9e 6 5 Ï } 9 Ï } e 6 5 3e 310. e x p 5e x 1 3 5 5e x 1 x 1 3 5 5e 2x 1 339. B; When x 5 0, y 5 0.5e 0.5(0) 5 0.5; When x 5 1,y 5 0.5e 0.5(1) ø 0.8240. C; When x 5 0, y 5 2e 0.5(0) 5 2; when x 5 1,y 5 2e 0.5(1) ø 3.3011. }3ee x 5 3e 1 2 x41. A; When x 5 0, y 5 e 0.5(0) 1 2 5 3; When x 5 22,y 5 e 0.5(22) 1 2 ø 2.3712. }4exe 5 4x 4ex 2 4x 5 4e 23x 5 }4 e 3x42.y43.y13. Ï 3} 8e 9x 5 Ï 3 } 38 Ï } e 9x 5 2e 3x14. }6e4x8e 5 }3 4 e4x 2 1115. C; (4e 2x ) 3 5 4 3 (e 2x ) 3 5 64e 6x21x14(27e16. D; Î} 13 x)21x}3e 7 x 23Domain: all real numbers Domain: all real numbers108e5 Î} 13 x}Range: y > 0 Range: y > 03e 7 x 2344.y45. y5 Ï }}36e 13 2 7 x 1 2 (23)5 Ï } 36e 6 x 4(0, 2)5 Ï } 36 Ï } e 6 Ï } x 41(0, 1) (1, 0.10)21xy 5 2e 23x5 6e 3 x 22x21y 5 2e17. The power 2 should have been applied to the 3 also.2 1(1, 20.90)(3e 5x ) 2 5 3 2 (e 5x ) 2 5 9e 10xDomain: all real numbers Domain: all real numbers18. The term 22x rather than 2x should have been subtractedfrom 6x.Range: y > 0 Range: y > 21Copyright © by McDougal Littell, a division of Houghton Mifflin Company.428Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.46.y47.y54. Let m 5 }ny 5 2.5e 20.5x 1 2r , so n 5 rm and }r n 5 }1 m .y 5 0.6e x9( 0,2 )A 5 P(1, 1.63)11 1 }n2 r 5 P 11 1 }m2 1 5 P F 11 1 }m2 1 G rt(1, 3.52)As n approaches 1`, m approaches 1`, and3(3, 1.63)5 (1, 1.52)( 0,2 )( 0,5 )11 1 }213 xm2 1 approximates e.1 y 5 2.5e 20.5x21x( 2,5 )21So, A 5 P 11 1 }y 5 0.6e n2 r approximates A 5 Pe rt as nx 2 230f (x)}g (x) 5 }2e23xe 5 24x 2e23x 2 (24x) 5 2e x 2010Domain: all real numbers Domain: all real numbers approaches 1`.Range: y > 2 Range: y > 055. When x 5 2002 2 1997 5 5: y 5 1.28e 1.31x48.1f(x) Domain: all real numbers y 5 1.28e 1.31(5) 5 1.28e 6.55 ø 895f(x) 5 e x 1 3 2 22Range: y > 22About 895 million camera phones were shipped globally1f(x) 5 e xin 2002.2256. When t 5 1999 2 1989 5 10: y 5 738e 0.345t(1, 1.36)y 5 738e 0.345(10) 5 738e 3.45 ø 23,2471 3 x( 0,2 )About 23,247 termites were collected in 1999.3( 23, 22 ) (22, 20.64)57. When P 5 2000, r 5 0.04, and t 5 5: A 5 Pe rtA 5 2000e (0.04 p 5) 5 2000e 0.2 ø 2442.8149.g(x) Domain: all real numbersThe balance after 5 years is $2442.81.(2, 4.62)Range: y > 1(1, 3.62)58. When P 5 800, r 5 0.0265 and t 5 12.5: A 5 Pe rtA 5 800e4g(x) 5 e x 2 1 1 1 73( 1,3 )5 800e 0.33125 ø 1114.17The balance after 12.5 years is $1114.17.44 3 x59. a. When k 5 20.02: L(x) 5 100eg(x) 5 e x 213 ( 0,3 )→ L(x) 5 100e 20.02xL(x)1009050.h(x)Domain: all real numbers80570Range: y > 2360(1, 0.14)50(0, 1)40h(x) 5 e 22x30(21, 22)(0, 22.86)005 x201020 40 60 80 xh(x) 5 e 22(x 1 1) 2 3Depth below water surface (m)51. Using the table feature, you can notice that incrementingsmall values of n increases the function very slowly.b. Using the graph, you can estimate that the percent ofsurface light is about 45% at a depth of 40 meters.Incrementing n by powers of 10 gives values that are onec. Using the graph, you can estimate that the submarinedigit closer to the actual value of e each time. Whenn 5 10 10 can descend about 35 meters before only 50% of, the value of the function is 2.718281828,surface light is available.which is the value of e correct to 9 decimal places.60. a. P(t) 5 P52. No, e cannot be expressed as a ratio of two integers0e 0.116t ; when P 05 30: P(t) 5 30e 0.116tbecause it is an irrational number; it is a decimal thatb. P(t)90neither terminates nor repeats.807053. Choose a, b, r, and q, such that a > 0, b > 0, r < 0, q < 0,60and r 2 q > 0.50Sample answer: f (x) 5 2e 23x , g (x) 5 e 24x4000 2 4 6 8 tHours after 1:00 P.M.Percent of lightPopulationc. Using the graph, you can estimate that the bacteriapopulation is about 48 at 5:00 p.m.Algebra 2Worked-Out Solution Key429
Chapter 7, continuedd. P(2.75) 5 30e 0.116(2.75) 5 30e 0.319 ø 41There are 2.75 hours between 1:00 p.m. and 3:45 p.m.so, let t 5 2.75.There are about 41 bacteria at 3:45 p.m.61. A 5 A 0e 20.05t where A 05 4 and t 5 14A 5 4e 20.05(14) 5 4e 20.7 ø 1.99The area after 14 days is about 2 square centimeters.62. a.69. f (x) 5 2x 70. f(x) 5 5x 2 3y 5 2x y 5 5x 2 3x 5 2y x 5 5y 2 3x}2 5 y x 1 3 5 5yf 21 (x) 5 x }2x 1 3} 5 y5f 21 (x) 5 x 1 3 }571. f (x) 5 24x 1 14 72. f (x) 5 1 }3 x 1 4X=0 Y=630y 5 24x 1 14 y 5 1 }3 x 1 4The arch is 630 feet tall at its highest point.b. The x-intercepts are approximately x 5 6315, so theends of the arch are about 315 2 (2315) 5 630 feetapart.Mixed Review63. ⏐x 1 8⏐ 5 13x 1 8 5 213 or x 1 8 5 13x 5 221 or x 5 564. ⏐3x 1 17⏐ 5 163x 1 17 5 216 or 3x 1 17 5 163x 5 233 or 3x 5 21x 5 211 or x 5 2 }1 365. 2x 2 2 4x 1 9 5 0x 5 4 6 Ï} 16 2 4(9 p 2)}} 5 4 6 Ï} 256}445 4 6 2i Ï} 14} 5 2 6 i Ï} 14}4266. x 2 1 12x 2 3 5 0x 5 212 6 Ï }}144 2 4(23 p 1)}} 5 212 6 Ï} 156}225 212 6 2 Ï} 39} 5 26 6 Ï } 39267. Ï } 5x 1 9 5 75x 1 9 5 495x 5 40x 5 868. Ï } 15x 1 34 5 x 1 615x 1 34 5 (x 1 6) 215x 1 34 5 x 2 1 12x 1 360 5 x 2 2 3x 1 20 5 (x 2 2)(x 2 1)x 5 2, x 5 1x 5 24y 1 14 x 5 1 }3 y 1 4x 2 14 5 24y x 2 4 5 }1 3 y14 2 x} 5 y43x 2 12 5 yf 21 (x) 5 }14 2 x4f 21 (x) 5 3x 2 1273. f (x) 5 212x 2 6 74. f (x) 5 2 1 }4 x 1 7y 5 212x 2 6 y 5 2 1 }4 x 1 7x 5 212y 2 6 x 5 2 1 }4 y 1 7x 1 6 5 212y x 2 7 5 2 }1 4 y2 }x 1 6 5 y1224x 1 28 5 yf 21 (x) 5 2 }x 1 61228 2 4x 5 yf 21 (x) 5 28 2 4xQuiz 7.1–7.3 (p. 498)1.y2.(1, 6) (3, 6)3(0, 2) (2, 2)y 5 2 ? 3 xy 5 2 ? 3 x 2 21xDomain: all real numbers Domain: all real numbersRange: y > 0 Range: y > 03.f(x) Domain: all real numbers1421,3Range: y > 28( 21,3)( )( )3 xf(x) 5 1 28(0, 3)(0, 1)1( )1 3 x xf(x) 58211yxCopyright © by McDougal Littell, a division of Houghton Mifflin Company.4. 3e 4 p e 3 5 3e 4 1 3 5 3e 7430Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.5. (25e 3x ) 3 5 25 3 (e 3x ) 3 5 2125e 9x6. }e4x5e 5 }1 5 e4x 2 1 7. }8e5x6e8.211yx9.5 4e5x 2 2x2x }2131y5 4e3x }3Domain: all real numbers Domain: all real numbersRange: y > 0 Range: y > 010.y11.g(x)2(1, 2.72)(0, 1)y 5 e x (0, 0.72)(21, 21) 1 xy 5 e x 1 1 2 2(0, 4)211(0, 5)g(x) 5 4e 23x 1 1(1, 1.20)(1, 0.20)xg(x) 5 4e 23xDomain: all real numbers Domain: all real numbersRange: y > 22 Range: y > 112. Decay factor 5 0.85;percent decrease 5 1 2 0.85 5 0.15nDomain: 0, 1, 2, 3, 430Range: 26.8, 22.8, 19.4,2016.5, 14.0; using thegraph you can estimate10that the number of0black-and-white TVs0 2 4 6 8 tsold in 1999 (t 5 2) wasYears since 1997about 19 million.13. When P 5 1200, r 5 0.045, and t 5 5: A 5 Pe rtA 5 1200e (0.045)(5) 5 1200e (0.225) ø 1502.79The balance after 5 years is $1502.79.TVs sold (millions)Lesson 7.47.4 Guided Practice (pp. 499 – 503)1. log 381 5 4 2. log 77 5 13 4 5 81 7 1 5 73. log 141 5 0 4. log 1/232 5 2514 0 5 1 1 }1 22 25 5 325. 2 5 5 32, so log 232 5 5 6. 27 1/3 5 3, so log 273 5 }1 37. log 12 ø 1.079 8. ln 0.75 ø 20.2889. s 5 93 log d 1 65 5 93 log 150 1 65ø 93(2.176) 1 65 5 267.368The wind speed near the tornado’s center is about267 miles per hour10. 8 log 8 x 5 x 11. log 77 23x 5 23x12. log 264 x 5 log 22 6x 5 6x 13. e ln 20 5 e log e 20 5 20x14. From the definition of logorithm, the inverseof y 5 4 x is y 5 log 4x.15. y 5 ln (x 2 5)x 5 ln ( y 2 5)e x 5 y 2 5e x 1 5 5 yThe inverse of y 5 ln (x 2 5) is y 5 e x 1 5.16. y11Domain: x > 0Range: all real numbers17. y y 5 log 1/3 (x 2 3)11( 9 , 2 )28( 9 , 2 )10( 3 , 1 )1( 3 , 1 )(4, 0)(1, 0)1y 5 log 1/3 xDomain: x > 3Range: all real numbers18. f(x)222(1, 0)(4, 1)(0, 22)(15, 0) x(3, 21)xf(x) 5 log 4 x(16, 2)f(x) 5 log 4 (x 1 1) 2 2Domain: x > 21Range: all real numbers7.4 Exercises (pp. 503 – 505)Skill Practice1. A logarithm with base 10 is called a(n) commonlogarithm.2. From the definition of logarithm, the inverse of y 5 5 x isy 5 log 5x, so y 5 5 x and y 5 log 5x are inverses.3. log 416 5 2 4. log 7343 5 34 2 5 16 7 3 5 34315. log 6 }36 5 22 6. log 64 1 5 06 22 5 }1 64 0 5 1367. The 23 and }1 8 should be switched. Because 223 5 }1 8 ,1the equation should be log 2 }8 5 23.8. 15 1 5 15, so log 1515 5 1.9. 7 2 5 49, so log 749 5 2.10. 6 3 5 216, so log 6216 5 3.xAlgebra 2Worked-Out Solution Key431
Chapter 7, continued11. 2 6 5 64, so log 264 5 6.45. y46.1yThe inverse of y 5 6 1 log x is y 5 10 x 2 612. 9 0 5 1, so log 91 5 0.11x13. 1 }1 22 23 21x5 8, so log 1/28 5 23.14. 3 23 5 }1 27 , so log 1}3 27 5 23.15. 16 21/2 5 }1 4 , so log 1}16 4 5 2 }1 2 .Domain: x > 0 Domain: x > 0Range: all real numbers Range: all real numbers16. 1 }1 42 22 47. y48. y5 16, so log 1/416 5 22.17. 8 3 5 512, so log 8512 5 3.1118. 5 4 5 625, so log 5625 5 4.21x1x19. 11 2 5 121, so log 11121 5 2.20. log 14 ø 1.146 21. ln 6 ø 1.792Domain: x > 0 Domain: x > 022. ln 0.43 ø 20.844 23. log 6.213 ø 0.793Range: all real numbers Range: all real numbers24. log 27 ø 1.431 25. ln 5.38 ø 1.68349. y26. log 0.746 ø 20.127 27. ln 110 ø 4.700y 5 log 2 x28. 7 log 7 x 5 x 29. log 55 x (4, 2)5 x(7, 2)30. 30 log 30 4 5 4 31. 10 log 8 15 8(2, 1) (5, 1)32. log 636 x 5 log 66 2x 21 (1, 0) (4, 0) x5 2x33. log 381 x 5 log 33 4x 5 4x34. log 5125 x 5 log 55 3x 5 3x35. log 232 x 5 log 22 5x 5 5xy 5 log 2 (x 2 3)Domain: x > 3Range: all real numbers36. B; log 100 x 5 log 1010 2x 5 2x50. y37. The inverse of y 5 log 8x is y 5 8 x .y 5 log 3 x 1 438. The inverse of y 5 7 x is y 5 log 7x.(3, 5)(9, 6)39. The inverse of y 5 (0.4) x (1, 4)is y 5 log 0.4x.(3, 1)40. The inverse of y 5 log 1/2x is y 5 1 }1 22 x 2(9, 2).22 (1, 0) y 5 log 3 x x41. y 5 e x 1 2x 5 e y 1 2Domain: x > 0Range: all real numbersln x 5 y 1 251.f(x) f(x) 5 log 4 x(16, 2)ln x 2 2 5 y(4, 1)The inverse of y 5 e x 1 2 is y 5 ln x 2 2.1(14, 1)42. y 5 2 x 22 (2, 0)x2 3(21, 21) (1, 0)x 5 2 y 2 3x 1 3 5 2 yf(x) 5 log 4 (x 1 2) 2 1log 2(x 1 3) 5 yDomain: x > 22The inverse of y 5 2 x 2 3 is y 5 log 2(x 1 3).Range: all real numbers43. y 5 ln (x 1 1)52. g(x) g(x) 5 log 6 (x 2 4) 1 2(40, 4)x 5 ln ( y 1 1)(5, 2)e x 5 y 1 12(10, 3)e x 2 1 5 y(1, 0)(6, 1)(36, 2)The inverse of y 5 ln (x 1 1) is y 5 e x 2 1.24g(x) 5 log 6 xx44. y 5 6 1 log xDomain: x > 4x 5 6 1 log yRange: all real numbersx 2 6 5 log y10 x 2 6 5 yCopyright © by McDougal Littell, a division of Houghton Mifflin Company.432Algebra 2Worked-Out Solution Key
Chapter 7, continued53. h(x)h(x) 5 log 5 x1(5, 1)21 (1, 0)x(4, 22)(0, 23)h(x) 5 log 5 (x 1 1) 2 3Domain: x > 21Range: all real numbers54. log 279 5 x 55. log 832 5 x27 x 5 9 8 x 5 323 3x 5 3 2 2 3x 5 2 53x 5 2 3x 5 5b. M 5 0.29(ln E ) 2 9.9M 1 9.9 5 0.29(ln E )M 1 9.9} 5 ln E0.29e (M 1 9.9)/0.29) 5 EThe inverse of the function, E 5 e (M 1 9.9)/0.29) ,represents the amount of energy released (in ergs) byan earthquake of magnitude M.62. a.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.x 5 2 }3x 5 5 }356. log 125625 5 x 57. log 4128 5 x125 x 5 625 4 x 5 1285 3x 5 5 4 2 2x 5 2 73x 5 4 2x 5 7x 5 4 }3P58. When P 5 57,000: h 5 28005 ln }101,300h 5 28005 ln 1 57,000 }101,3002 ø 4603x 5 7 }2The altitude is about 4603 meters above sealevel when the air pressure is 57,000 pascals.59. When H 1 5 10 22.3 : pH 5 2log[H1]pH 5 2log[10 22.3 ] 5 2.360.The pH of lemon juice is 2.3.Length (in.)l1401201008060402000 40 80 120 160 200 240 280 wWeight (lb)12 in.10 ft p } 5 120 in.1 ftUsing the graph, you can estimate that the alligatorweighs about 281 pounds when it is 10 feet (120 inches)long.61. a. When E 5 2.5 3 10 24 : M 5 0.29(ln E) 2 9.9M 5 0.29 ln (2.5310 24 ) 2 9.9ø 0.29(56.1783) 2 9.9 ø 6.39The magnitude of the earthquake was about 6.4.b. Using the graph, you can estimate that there are about15 different kinds of fish species when the area is30,000 square meters.c. Using the graph, you can estimate that the area of thelake is about 4000 square meters when there are 6species of fish.d. As the area of the pool or lake increases, the numberof fish species also increases. The answer makes sensebecause the larger the pool or lake, the more roomthere is for more varieties of fish to thrive.63. s 5 0.159 1 0.118(log d )s 2 0.159 5 0.118(log d )s 2 0.159} 5 log0.118 10ds 2 0.159}10 0.118 5 dThe inverse of the function is d 5 10 (s 2 0.159)/0.118 .When s 5 0.2: d 5 10 1 }0.2 2 0.159 20.118 ø 2.23Sand particles on a beach with a slope of 0.2 have anaverage diameter of about 2.23 millimeters.Mixed Review64. 2 3 p 2 5 5 2 3 1 5 5 2 8 5 25665. (5 23 ) 2 5 5 23 p 2 5 5 26 5 }1 5 5 16 }15,62566. 8 1 p 8 3 p 8 25 5 8 1 1 3 2 5 5 8 21 5 }1 81 } 167. 1 }5 32 23 5 }5233 5 5 3 }12523 } 5 } 5 }27 1 1 125}3 3 }2768. }10610 5 4 106 2 4 5 10 2 5 10069. (6 22 ) 21 5 6 22 p 21 5 6 2 5 3670. }424 5 5 42 2 5 5 4 23 5 }1 4 5 1 3 }6471. 1 }7872 22 5 78(22) 72169 }9(22)5 }7 7 5 218 7216 2 (218) 5 7 2 5 49Algebra 2Worked-Out Solution Key433
Chapter 7, continued72. x 1/2 p x 2/3 5 x 1/2 1 2/3 5 x 7/673. (m 9 ) 21/6 5 m 29/6 5 m 23/2 15 }m 5 Ï} m3/2 }m 274. 3 Ï } 54x 6 y 3 5 3 Ï } 27x 6 y 3 3 Ï } 2 5 3x 2 y 3 Ï } 275. (n 4/3 p n 2/5 ) 1/6 5 (n 4/3 1 2/5 ) 1/6 5 (n 26/15 ) 1/65 n 26/15 p 1/6 5 n 26/90 5 n 13/4576. x1/4 y 3}x 5/2 y 5 1/2 x1/4 2 5/2 y 3 2 1/2 5 x 29/4 y 5/2 5 y5/2}x 5 x3/4 y 5/29/4 }x 3Î}x1677. 4 }y 5 Ï 4 } x 1612 4Ï } y5 }x412 y 378. 1 5 Ï } x 10 p 3 Ï } x 9 2 2 5 (x 2 p x 3 ) 2 5 (x 2 1 3 ) 2 5 (x 5 ) 2 5 x 1079. 5 Ï} x p Ï } x 73Ï 5 5x1/2 p x } }7/2250x 16 (250x 16 )52 16/3x45 x24/3 }3}3Ï } 21 7/25x1/21/35 }Ï 5 1} 2 x 4/3 Ï 5 Ï 3 } 4x 3 } }22 2x 2250 1/3 x 5 5x 416/3 5 Ï 3 } 2 x 16/3b. R(n) 5 2 n ; A(n) 5 1 }2 n ; R(n) represents exponentialgrowth and A(n) represents exponential decay.3. Sample answer:Let r 5 0.05.Item of increasing value: Item of decreasing value:y 5 a(1 1 0.05) ty 5 a(1 2 0.05) t100 5 a(1.05) 2 100 5 a(0.95) 290.70 ø a 110.80 ø aAn equation isAn equation isy 5 90.70(1.05) t . y 5 110.80(0.95) t .Value (dollars)y120906030(2, 100)y 5 90.70(1.05) ty 5 110.80(0.95) tMixed Review of Problem Solving (p. 506)1. a. Let y 5 log 3(x 2 h) 1 k.When x 5 3; y 5 log 3(3 2 h) 1 k 5 1When x 5 5; y 5 log 3(5 2 h) 1 k 5 2log 3(5 2 h) 2 log 3(3 2 h) 5 1When h 5 2:log 3(3 2 h) 1 k 5 1log 3(3 2 2) 1 k 5 1log 31 1 k 5 10 1 k 5 1log 5 2 h }3 2 h 5 1 → h 5 2k 5 1Therefore, y 5 log 3(x 2 2) 1 1.b. y 5 log 3(x 2 2) 1 1x 5 log 3( y 2 2) 1 1x 2 1 5 log 3( y 2 2)3 x21 log3( y 2 2)5 33 x21 5 y 2 23 x 2 1 1 2 5 y1y12. a. Fold number 0 1 2 3 4Number of regions 1 2 4 8 16Fractional area ofeach regionx1 1 }21}41}81}1600 2 4 6 8 tYears4. A 5 Pe rt when P 5 2000, r 5 0.04, and t 5 2:A 5 2000e 0.04(2) 5 2000e 0.08 ø 2166.57Your balance after 2 years is $2166.57.A 5 Pe rt when P 5 2000, r 5 0.04, and A 5 2250:2250 5 2000e 0.04t1.125 5 e 0.04tln(1.125) 5 0.04t0.1177 ø 0.04t2.94 ø tAfter 3 full years for your balance will exceed $2250.You must round up because the answer must be in fullyears.5. a. A 5 P 11 1 }n2r ntFirst CD: t 5 3, p 5 1500, r 5 0.02 and n 5 12A 5 150011 1 }0.0212 2 12(3) 5 15001 }12.0212 2 36 ø 1592.68Second CD: t 5 5, p 5 2000, r 5 0.03, and n 5 12A 5 2000 1 1 1 0.03 }12 2 12(5) 5 2000(1.0025) 60 ø 2323.23The first CD yields $1592.68 at the end of its term andearns 1592.68 2 1500 5 $92.68 interest. The secondCD yields $2323.23 at the end of its term and earns2323.23 2 2000 5 $323.23 interest.b. The difference in the amounts of interestis $323.23 2 $92.68 5 $230.55.c. Sample answer: The first CD requires a smallminimum balance, and the money can be withdrawnsooner, but the second CD earns more interest.6. y 5 10e 20.0564t when t 5 10:y 5 10e 20.0564(10) 5 10e 20.564 ø 5.69There are about 5.69 milligrams of tritium left after 10years.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.434Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.7. a.Oil collected(billions of barrels)y60504030201000 500 1000 1500 2000 2500 xWells drilledb. Using the graph, you can estimate that you couldexpect to collect about 40 billion barrels after drilling1000 wells.c. Using the graph, you can estimate that about 2400wells need to be drilled in order to collect 50 billionbarrels of oil.Lesson 7.57.5 Guided Practice (pp. 507–509)1. log6 }5 8 5 log 6 5 2 log 8 ø 0.898 2 1.161 5 20.26362. log 640 5 log 6(8 p 5) 5 log 68 1 log 65ø 1.161 1 0.898 5 2.0593. log 664 5 log 68 2 5 2log 68 ø 2(1.161) 5 2.3224. log 6125 5 log 65 3 5 3log 65 ø 3(0.898) 5 2.6945. log 3x 4 5 log 3 1 log x 4 5 log 3 1 4log x6. ln 4 1 3 ln 3 2 ln 12 5 ln 4 1 ln 3 3 2 ln 125 ln 4 1 ln 27 2 ln 125 ln }4 p 27 5 ln }10812 12 5 ln 97. log 58 5 }log 8log 5 ø }0.90310.6990 ø 1.292log 148. log 814 5 }log 8 ø }1.14610.9031 ø 1.2699. log 269 5 }log 9log 26 ø }0.95421.4150 ø 0.674log 3010. log 1230 5 }log 12 ø }1.47711.0792 ø 1.36911. Increase in loudness 5 L(3I ) 2 L(I )5 10 log }3I 2 10 log }I I 0I 05 10 1 log 3I }I 02 log I }I 025 10 1 log 3 1 log I }I 02 log I }I 025 10(log 3) ø 4.8The loudness increases by about 4.8 decibels.7.5 Exercises (pp. 510 – 513)Skill Practice1. To condense the expression log 32x 1 log 3y, youneed to use the product property of logarithms.2. You can evaluate log 712 by using thechange-of-base formula with common logarithmsor with natural logarithms, and then using acalculator to evaluate the result.3. B; ln 6 2 ln 2 5 ln 6 }2 5 ln 34. D; 2 ln 6 5 ln 6 2 5 ln 365. A; 6 ln 2 5 ln 2 6 5 ln 646. C; ln 6 1 ln 2 5 ln 6 p 2 5 ln 127. log 3 5 log }12 5 log 12 2 log 44ø 1.079 2 0.602 5 0.4778. log 48 5 log 12 p 4 5 log 12 1 log 4ø 1.079 2 0.602 5 1.6819. log 16 5 log 4 2 5 2 log 4 ø 2(0.602) 5 1.20410. log 64 5 log 4 3 5 3 log 4 ø 3(0.602) 5 1.80611. log 144 5 log 12 2 5 2 log 12 ø 2(1.079) 5 2.15812. log }1 3 5 log }4 5 log 4 2 log 1212ø 0.602 2 1.079 5 20.47713. log }1 5 log 1 2 log 4 ø 0 2 0.602 5 20.602414. log }1 5 log 1 2 log 12 ø 0 2 1.079 5 21.0791215. log 34x 5 log 34 1 log 3x16. ln 15x 5 ln 15 1 ln x17. log 3x 4 5 log 3 1 log x 4 5 log 3 1 4 log x18. log 5x 5 5 5 log 5x19. log 22 }5 5 log 2 2 2 log 2 520. ln }12 5 ln 12 2 ln 5521. log 4x }3y 5 log 4 x 2 log 4 3y 5 log 4 x 2 log 4 3 2 log 4 y22. ln 4x 2 y 5 ln 4 1 ln x 2 1 ln y 5 ln 4 1 2 ln x 1 ln y23. log 75x 3 yz 2 5 log 75 1 log 7x 3 1 log 7y 1 log 7z 25 log 75 1 3 log 7x 1 log 7y 1 2 log 7z24. log 636x 2 5 log 636 1 log 6x 2 5 log 636 1 2 log 6x25. ln x 2 y 1/3 5 ln x 2 1 ln y 1/3 5 2 ln x 1 1 }3 ln y26. log 10x 3 5 log 10 1 log x 3 5 log 10 1 3 log x27. log 2Ï } x 5 log 2x 1/2 5 1 }2 log 2 x28. ln }6x2y 5 ln 4 6x2 2 ln y 4 5 ln 6 1 ln x 2 2 ln y 45 ln 6 1 2 ln x 2 4 ln y29. ln 4 Ï } x 3 5 ln x 3/4 5 3 }4 ln x30. log 3Ï } 9x 5 log 33 Ï } x 5 log 33x 1/2 5 log 33 1 log 3x 1/25 log 33 1 1 }2 log 3 x31. When expanding the logarithmic expression, thequantities (log 25) and (log 2x) should have been added,not multiplied.log 25x 5 log 25 1 log 2xAlgebra 2Worked-Out Solution Key435
Chapter 7, continuedlog 1732. When expanding the logarithmic expression, the 3 should 51. log 617 5 }log 6 ø }1.23040.7782 ø 1.581have been multiplied by ln x, not ln 8.ln 8x 3 5 ln 8 1 ln x 3 log 285 ln 8 1 3 ln x52. log 228 5 }log 2 ø }1.44720.3010 ø 4.808733. log 47 2 log 410 5 log 4 }log 191053. log 719 5 }log 7 ø }1.27880.8451 ø 1.51334. ln 12 2 ln 4 5 ln }12 4 5 ln 3log 4854. log 448 5 }log 4 ø }1.68120.6021 ø 2.79235. 2 log x 1 log 11 5 log x 2 1 log 11 5 log 11x 2log 2736. 6 ln x 1 4 ln y 5 ln x 6 1 ln y 4 5 ln x 6 y 455. log 927 5 }log 9 ø }1.43140.9542 ø 1.50037. 5 log x 2 4 log y 5 log x 5 2 log y 4 5 log }x5log 32y 456. log 832 5 }log 8 ø }1.50510.9031 ø 1.66738. 5 log 42 1 7 log 4x 1 4 log 4y5 log 42 5 1 log 4x 7 1 log 4y 457. log 24 log }246}5 5 5 log 24 2 log 5} 5 }}log 6 log 65 log 432x 7 y 41.3802 2 0.6990ø }} ø 0.87539. ln 40 1 2 ln }1 2 1 ln x 5 ln 40 1 ln 1 }1 22 2 0.77821 ln xlog }15155 ln 40 1 ln }1 58. log4 1 ln x2}7 5 7 log 15 2 log 7} 5 }}log 2 log 25 ln 401 }1 1.1761 2 0.8451ø }} ø 1.10042 x0.30105 ln 10x40. log 54 1 }1 59. log 9 log }93}3 log 5 x 5 log 5 4 1 log 40 5 40 log 9 2 log 40} 5 }}log 3 log 35 x1/35 log 54x 1/3 5 log 54 Ï 3 } x0.9542 2 1.6021ø }} ø 21.35841. 6 ln 2 2 4 ln y 5 ln 2 6 2 ln y 4 5 ln 64 2 ln y 4 5 ln }64 0.4771y 4log }3 360. log42. 2(log 320 2 log 34) 1 0.5 log 347 }16 5 16 log 3 2 log 16} 5 }}log 7 log 75 21 log 20 3}4 2 1 0.5 log 340.4771 2 1.2041ø }} ø 20.8600.84515 2 log 35 1 0.5 log 3461. When using the change-of-base formula, log 7 should5 log 35 2 1 log 34 0.5have been divided by log3.5 log 325 1 log 32log 37 5 }log 75 log 325 p 2log 35 log 35062. a. I 5 10 24 w/m 2 ; I 05 10 212 w/m 243. C; 3 log 46 5 log 46 3 5 log 4216L(I ) 5 10 log }I 5 10 log44. D; log 38 1 2 log 33 5 log 38 1 log 33 2I 01 }1024102 5 10 log 212 1085 10 p 8 5 805 log 38 1 log 39 5 log 38 p 95 log 372The dog produces about 80 decibels of sound.log 372 Þ log 348b. I 5 10 0 w/m 2 ; I 05 10 212 w/m 245. log 47 5 }log 7log 4 ø }0.84510.6021 ø 1.404L(I ) 5 10 log }I 5 10 logI 0 1}108102 212log 1346. log 513 5 }log 5 ø }1.11390.6990 5 1.5945 10 log 10 12 5 10 p 12 5 120log 1547. log 315 5 }log 3 ø }1.1761The ambulance siren produces about 120 decibels of0.4771 ø 2.465sound.log 2248. log 322 5 }log 8 ø }1.3424c. I 5 10 26.5 w/m 2 ; I 05 10 212 w/m 20.9031 ø 1.486L(I ) 5 10 log }I 5 10 log49. log 36 5 }log 6log 3 ø }0.7782I 01 }1026.5102 5 10 log 212 105.50.4771 ø 1.6315 10 p 5.5 5 55log 1450. log 514 5 }log 5 ø }1.14610.6990 ø 1.640 The bee produces about 55 decibels of sound.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.436Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.63. I 5 10 3 w/m 2 ; I 05 10 212 w/m 2L(I ) 5 10 log }I 5 10 logI 01 }103102 5 10 log 212 10155 10 p 15 5 150The trumpet produces about 150 decibels of sound.64. a. Sample answer: The statement log b(M 1 N) 5log bM 1 log bN is false when b 5 2, M 5 3, and N 5 4.log 2(3 1 4) 0 log 23 1 log 24log 27 0 log 23 1 log 24log 27 0 log 23 p 4log 27 Þ log 212b. Sample answer: The statement log b(M 2 N) 5log bM 2 log bN is false when b 5 2, M 5 4, and N 5 3log 2(4 2 3) 0 log 24 2 log 23log 2(1) 0 log 24 2 log 23log 21 Þ log 24 }365. Let x 5 log bm and y 5 log bn. Then m 5 b x and n 5 b y .log bmn 5 log b(b x b y ) 5 log b(b x 1 y ) 5 x 1 y5 log bm 1 log bn66. Let x 5 log bm and y 5 log bn. Then m 5 b x and n 5 b y .b xmlog b } n 5 log b }b y 5 log b bx 2 y 5 x 2 y 5 log bm 2 log bn67. Let x 5 log bm. Then m 5 b x and m n 5 b nx .log bm n 5 log bb nx 5 nx 5 n log bm68. Let x 5 log ba, y 5 log bc, and z 5 log ca.Then a 5 b x , c 5 b y , and a 5 c z , so that b x 5 c z .x 5 log ba 5 log bb x 5 log bc z 5 zlog bc 5 zyx 5 zy → z 5 x } ylog ca 5 log b a}log bcProblem Solving69. I 5 1.4 3 10 25 w/m 2 ; I 05 10 212 w/m 2 ; n 5 3L(I ) 5 10 log }1 nII 02 5 10 log 3 3 1.4 3 10251 }}1022125 10 log 42,000,000 ø 10(7.62325) ø 76.2325The decibel level of the combined conversations in theroom is about 76 decibels.70. I 5 3.2 3 10 24 w/m 2 ; I 05 10 212 w/m 2 ; n 5 5L(I ) 5 10 log }1 nII 02 5 10 log 5 3 3.2 3 10241 }}1022125 10 log (1,600,000,000) ø 10(9.204) ø 92.04All five cars in the parking garage makeabout 92 decibels of sound.71. Increase in loudness5 L(10I ) 2 L(I )5 10 log 10I }I 02 10 log I }I 05 101log 10I }I 02 log I }I 025 101log 10 1 log I }I 02 log I }I 02Substitute.Distributive propertyProduct Property5 10 log 10 Simplify.5 10 log 1010 1 5 1The loudness increases by 10 decibels.72. Blue whale: I 5 10 6.8 w/m 2 ; I 05 10 212 w/m 2L(I ) 5 10 log1 I }I 02 5 10 log 1 }106.810 212 2ø 10 log (10 18.8 ) 5 10 p 18.8 5 188Human: I 5 10 0.8 w/m 2 ; I 05 10 212 w/m 2L(I ) 5 10log1 I }I 02 5 10log 1 }100.810 212 2 ø 10log(1012.8 )5 10 p 12.8 5 128A blue whale can produce sounds that are188 2 128 5 60 decibels louder than thoseproduced by a human.73. a. s 5 log 2f 2 5 2log 2fb. f 1.414 2.000 2.828 4.000s 1 2 3 4f 5.657 8.000 11.314 16.000s 5 6 7 8The amount of light that enters the camera increasesby about 1 each time.c. When s 5 9, an equation relating s and f is:9 5 2 log 2f9}2 5 log 2 f2 9/2 5 2 log 2 f2 9/2 5 fSo, the ninth f -stop is about 2 9/2 ø 22.627.74. a. Let h 5 original altitude, so that 2h is the doubledaltitude.Increase in speed 5 s(2h) 2 s(h)5 2 ln[100(2h)] 2 2ln (100h)5 2[ln (200h) 2 ln (100h)]5 2[ln (2 p 100h) 2 ln (100h)]5 2[ln 2 1 ln (100h) 2 ln (100h)]5 2 ln 2ø 1.39The wind speed increases by about 1.39 knots whenthe altitude doubles.Algebra 2Worked-Out Solution Key437
Chapter 7, continuedlog (100h)b. s(h) 5 2 ln (100h) 5 2 1}log e 2 5 2} p log (100h)log e25 }log e (log 100 1 log h) 5 2} (2 1 log h)log eMixed Review75.F 5 2812 20G F 1 6 722 219G F 5 11 2110 1G27 1176.F6 3G F 2 29 17213 1G F 5 2 2619 2G1.7 2.4 6.877. 3F9.2 5.3 7.2G F 5 5.1 7.2 20.427.6 15.9 21.6G78. Ï } x 1 12 1 4 5 11 79. Ï 3} x 1 10 1 6 5 4Ï } x 1 12 5 73Ï } x 1 10 5 22Check:x 1 12 5 49 x 1 10 5 28Ï } x 1 12 1 4 5 11Ï } 37 1 12 1 4 0 11x 5 37 x 5 218Ï } 49 1 4 0 11Check:3Ï } x 1 10 1 6 5 43Ï } 218 1 10 1 6 0 43Ï } 28 1 6 0 47 1 4 0 11 22 1 6 0 411 5 11 ✓ 4 5 4 ✓37 is the solution. 218 is the solution.80. Ï } x 1 6 5 Ï } 3x 2 14 81. 3 Ï } 2x 2 7 5 3 Ï } 8 2 xx 1 6 5 3x 2 142x 2 7 5 8 2 x22x 5 220 3x 5 15x 5 10 x 5 5Check:Check:Ï } x 1 6 5 Ï } 3x 2 143Ï } 2x 2 7 5 Ï 3} 8 2 xÏ } 10 1 6 0 Ï } 3(10) 2 143Ï } 2(5) 2 7 0 Ï 3} 8 2 5Ï } 16 0 Ï } 30 2 143Ï } 10 2 7 0 Ï 3 } 3Ï } 16 5 Ï } 16 ✓3Ï } 3 5 Ï 3 } 3 ✓10 is the solution. 5 is the solution.82. Ï } x 2 1 5 x 2 3x 2 1 5 x 2 2 6x 1 90 5 x 2 2 7x 1 100 5 (x 2 5)(x 2 2)x 5 5 or x 5 2Check x 5 5: Check x 5 2:Ï } 5 2 1 0 5 2 3 Ï } 2 2 1 0 2 2 3Ï } 4 0 2 Ï } 1 0 212 5 2 ✓ 1 Þ 212 is an extraneous solution, so the 5 is the only solution.83. x 1 2 5 Ï } 9x 1 28x 2 1 4x 1 4 5 9x 1 28x 2 2 5x 2 24 5 0(x 2 8)(x 1 3) 5 0x 5 8 or x 5 23Check x 5 8: Check x 5 238 1 2 0 Ï } 9 p 8 1 28 23 1 2 0Ï } 9(23) 1 2810 0 Ï } 100 21 0 Ï } 110 5 10 ✓ 21 Þ 123 is an extraneous solution, so 8 is the only solution.84. e 8 ø 2981 85. e 26 ø 0.0024886. e 3.5 ø 33.12 87. e 20.4 ø 0.67088. log 12 ø 1.079 89. log 1.8 ø 0.25590. ln 24 ø 3.178 91. ln 8.49 ø 2.139Quiz 7.4 – 7.5 (p. 513)1. 4 2 5 16, so log 416 5 2 2. 5 0 5 1, so log 51 5 03. 8 1 5 8, so log 88 5 14. 1 }1 22 25 5 32, so log 1/232 5 255.7.211yx6.y(1, 2)y 5 ln x 1 2(2, 2.69)1(2, 0.69)21 (1, 0)y 5 ln xDomain: x > 0 Domain: x > 0Range: all real numbersy(9, 2)2(3, 1)(21, 0) (5, 1)24 (1, 0)(23, 21)y 5 log 3 (x 1 4) 2 1Domain: x > 24y 5 log 3 xRange: all real numbers8. log 25x 5 log 25 1 log 2x9. log 5x 7 5 7log 5xxRange: all real numbers10. ln 5xy 3 5 ln 5 1 ln x 1 ln y 3 5 ln 5 1 ln x 1 3 ln y11. log 36y 4}x 8 5 log 3 6y4 2 log 3x 85 log 36 1 log 3y 4 2 log 3x 85 log 36 1 4 log 3y 2 8 log 3x512. log 35 2 log 320 5 log 3 }20 5 log 1}3 413. ln 6 1 ln 4x 5 ln 6 p 4x 5 ln 24x14. log 65 1 3 log 62 5 log 65 1 log 62 35 log 65 1 log 68 5 log 68 p 55 log 640xCopyright © by McDougal Littell, a division of Houghton Mifflin Company.438Algebra 2Worked-Out Solution Key
Chapter 7, continued15. 4 ln x 2 5 ln x 5 ln x 4 2 ln x 5 5 ln }x4x 5 ln 1 5 } xlog 1016. log 310 5 }log 3 ø 1}0.4771 ø 2.096log 1417. log 714 5 }log 7 ø }1.14610.8451 ø 1.356log 2418. log 524 5 }log 5 ø }1.38020.6990 ø 1.975log 4019. log 840 5 }log 8 ø }1.60210.9031 ø 1.77420. I 5 10 24 W/m 2 ; I 05 10 212 W/m 2L(I ) 5 10 log1 I }I 02 5 10 log 1 }102410 212 25 10 log 10 8 5 10 p 8 5 80The alarm clock’s loudness is 80 decibels.Graphing Calculator Activity 7.5 (p. 514)1. 2.3. 4.13. The natural log, or ln x, is equal to log ex. So, if you didnot have a natural logarithm then you could graphy 5 ln x by graphing y 5 }log xlog e .Lesson 7.67.6 Guided Practice (pp. 515 – 519)1. 9 2x 5 27 x 2 1 2. 100 7x 1 1 5 1000 3x 2 2(3 2 ) 2x 5 (3 3 ) x 2 1 (10 2 ) 7x 1 1 5 (10 3 ) 3x 2 23 4x 5 3 3x 2 3 10 14x 1 2 5 10 9x 2 64x 5 3x 2 3 14x 1 2 5 9x 2 6x 5 23 5x 5 283. 81 3 2 x 5 1 }1 32 5x 2 6 4. 2 x 5 5x 5 2 8 }5(3 4 ) 3 2 x 5 (3 21 ) 5x 2 6 log 22 x 5 log 253 12 2 4x 5 3 25x 1 6 x 5 log 2512 2 4x 5 25x 1 6 x 5 log5 }log2 ø 2.32x 5 265. 7 9x 5 15 6. 4e 20.3x 2 7 5 13log 77 9x 5 log 715 4e 20.3x 5 209x 5 log 715 e 20.3x 5 5x 5 log15 }9log7 ø 0.15 ln e20.3x 5 ln 520.3x ø 1.6094Copyright © by McDougal Littell, a division of Houghton Mifflin Company.5. 6.7. 8.9. 10.11. 12.x ø 25.367. ln (7x 2 4) 5 ln (2x 1 11)7x 2 4 5 2x 1 115x 5 15x 5 3Check: ln (7x 2 4) 5 ln (2x 1 11)ln (7 p 3 2 4) 0 ln (2 p 3 1 11)ln 17 5 ln 17 ✓8. log 2(x 2 6) 5 52 log 2 (x 2 6) 5 2 5x 2 6 5 32x 5 38Check: log 2(x 2 6) 5 log 2(38 2 6) 5 log 232Because 2 5 5 32, log 232 5 5. ✓9. log 5x 1 log (x 2 1) 5 2log [5x(x 2 1)] 5 210 log[5x(x 2 1)] 5 10 25x(x 2 1) 5 1005x 2 2 5x 2 100 5 0x 2 2 x 2 20 5 0(x 2 5)(x 1 4) 5 0x 5 5 or x 5 24Algebra 2Worked-Out Solution Key439
Chapter 7, continuedCheck: log 5x 1 log (x 2 1) 5 2log 5 p 5 1 log (5 2 1) 0 2log 25 1 log 4 0 2log 100 0 22 5 2 ✓So, 5 is a solution.log 5x 1 log (x 2 1) 5 2log 5 p (24) 1 log ((24) 2 1) 0 2log (220) 1 log (25) 0 2Because log (220) and log (25) are not defined, 24 isnot a solution.10. log 4(x 1 12) 1 log 4x 5 3log 4[x(x 1 12)] 5 34 log 4 [x(x 1 12)] 5 4 3x(x 1 12) 5 64x 2 1 12x 5 64x 2 1 12x 2 64 5 0(x 1 16)(x 2 4) 5 0x 5 216 or x 5 4Check: log 4(x 1 12) 1 log 4x 5 3log 4(216 1 12) 1 log 4(216) 0 3log 4(24) 1 log 4(216) 0 3Because log 4(24) and log 4(216) are not defined, 24 isnot a solution.log 4(x 1 12) 1 log 4x 5 3log 4(4 1 12) 1 log 44 0 3log 416 1 log 44 0 3log 464 0 3Because 4 3 5 64, log 464 5 3. ✓So, 4 is a solution.11. M 5 5 log D 1 2 When M 5 77 5 5 log D 1 25 5 5 log D1 5 log D10 1 5 10 log D10 5 DThe diameter is 10 millimeters.7.6 Exercises (pp. 519 – 522)Skill Practice1. The equation 5 x 5 8 is an example of an exponentialequation.2. Sample answer: Logarithmic equations have extraneoussolutions when you have to factor to solve them.3. 5 x 2 4 5 25 x 2 6 4. 7 3x 1 4 5 49 2x 1 15 x 2 4 5 (5 2 ) x 2 6 7 3x 1 4 5 (7 2 ) 2x 1 15 x 2 4 5 5 2x 2 12 7 3x 1 4 5 7 4x 1 2x 2 4 5 2x 2 12 3x 1 4 5 4x 1 28 5 x 2 5 x5. 8 x 2 1 5 32 3x 2 2 6. 27 4x 2 1 5 9 3x 1 8(2 3 ) x 2 1 5 (2 5 ) 3x 2 2 (3 3 ) 4x 2 1 5 (3 2 ) 3x 1 82 3x 2 3 5 2 15x 2 10 3 12x 2 3 6x 1 165 33x 2 3 5 15x 2 10 12x 2 3 5 6x 1 167 5 12x 6x 5 197}12 5 x x 5 }19 67. 4 2x 2 5 5 64 3x 8. 3 3x 2 7 12 2 3x5 81(2 2 ) 2x 2 5 5 (2 6 ) 3x 3 3x 2 7 5 (3 4 )12 2 3x2 4x 2 10 5 2 18x 3 3x 2 7 5 3 48212x4x 2 10 5 18x3x 2 7 5 48 2 12x210 5 14x 15x 5 552 5 }7 5 x x 5 11 }39. 36 5x 1 2 5 1 1 }62 11 2 x 10. 10 3x 2 10 5 1 1 }1002 6x 2 1(6 2 ) 5x 1 2 5 (6 21 ) 11 2 x 10 3x 2 10 5 (10 22 ) 6x 2 16 10x 1 4 5 6 211 1 x 10 3x 2 10 5 10 212x 1 210x 1 4 5 211 1 x 3x 2 10 5 212x 1 29x 5 215 15x 5 12x 5 2 5 }3x 5 4 }511. 25 10x 1 8 5 1 }1252 1 12. 8 x 5 20(5 2 ) 10x 1 8 5 (5 23 ) 4 2 2x log 88 x 5 log 8205 20x 1 16 5 5 212 1 6x x 5 log 82020x 1 16 5 212 1 6xlog 20x 5 }log 814x 5 228 ø 1.44x 5 2213. e 2x 5 5 14. 7 3x 5 18ln e 2x 5 ln 5 log 77 3x 5 log 7182x 5 ln 5 3x 5 log 7182x ø 1.61log 18x 5 }3 log 7x ø 21.61 ø 0.5015. 11 5x 5 33 16. 7 6x 5 12log 1111 5x 5 log 1133 log 77 6x 5 log 7125x 5 log 1133 6x 5 log 712log 33x 5 }5 log 11log 12x 5 }6 log 7 ø 0.21ø 0.2917. 4e 22x 5 17 18. 10 3x 1 4 5 9e 22x 5 17 }4ln e 22x 5 ln 17 }422x 5 ln 17 }4x 5 2 1 }2 ln 17 }410 3x 5 5log 10 3x 5 log 53x 5 log 5x 5 1 }3 log 5x ø 20.72 x ø 0.23Copyright © by McDougal Littell, a division of Houghton Mifflin Company.440Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.19. 23e 2x 1 16 5 5 20. 0.5 x 2 0.25 5 423e 2x 5 211 0.5 x 5 4.25e 2x 5 11 }3ln e 2x 5 ln 11 }32x 5 ln 11 }3x 5 1 }2 ln 11 }3x ø 0.65log 0.50.5 x 5 log 0.54.2521. 1 }3 (6)24x 1 1 5 6 22. 2 0.1x 2 5 5 71}3 (6)24x 5 5 2 0.1x 5 12x 5 log 0.54.25log 4.25x 5 }log 0.5x ø 22.096 24x 5 15 log 22 0.1x 5 log 212log 66 24x 5 log 615 0.1x 5 log 21223. 3 }4 e2x 1 7 }2 5 4log 1224x 5 log 615 x 5 }0.1 log23}4 e2x 5 }1 2e 2x 5 2 }3ln e 2x 5 ln 2 }3log 15x 5 }24 log 6x ø 20.382x 5 ln 2 }3x 5 1 }2 ln 2 }3x ø 20.2024. log 5(5x 1 9) 5 log 56x5x 1 9 5 6x9 5 xCheck: log 5(5x 1 9) 5 log 56xlog 5(5 p (9 1 9)) 0 log 56 p 9log 554 5 log 554 ✓25. ln (4x 2 7) 5 ln (x 1 11)4x 2 7 5 x 1 113x 5 18x 5 6Check: ln (4x 2 7) 5 ln (x 1 11)ln (4 p 6 2 7) 0 ln (6 1 11)ln 17 5 ln 17 ✓x ø 35.826. ln (x 1 19) 5 ln (7x 2 8)x 1 19 5 7x 2 827 5 6x9}2 5 xCheck: ln (x 1 19) 5 ln (7x 2 8)ln 1 9 }2 1 19 2 0 ln 1 7 p 9 }2 2 8 2ln 47 }2 5 ln 47 }2 ✓27. log 5(2x 2 7) 5 log 5(3x 2 9)2x 2 7 5 3x 2 92 5 xCheck: log 5(2x 2 7) 5 log 5(3x 2 9)log 5(2 p 2 2 7) 0 log 5(3 p 2 2 9)log 5(23) 0 log 5(23)Because log 5(23) is not defined, 2 is not a solution.So, there is no solution.28. log (12x 2 11) 5 log (3x 1 13)12x 2 11 5 3x 1 139x 5 24x 5 8 }3Check: log (12x 2 11) 5 log (3x 1 13)log 1 12 p 8 }3 2 11 20 log 1 3 p 8 }3 1 13 2The solution is 8 }3 .log 21 5 log 21 ✓29. log 3(18x 1 7) 5 log 3(3x 1 38)18x 1 7 5 3x 1 3815x 5 31x 5 31 }15Check: log 3(18x 1 7) 5 log 3(3x 1 38)log 3 1 18 p 31 }15 1 7 20 log 3 1 3 p 31 }log 3 221 }The solution is 31 }15 .15 1 38 25 5 log }2213 5 ✓30. log 6(3x 2 10) 5 log 6(14 2 5x)3x 2 10 5 14 2 5x8x 5 24x 5 3Check: log 6(3x 2 10) 5 log 6(14 2 5x)log 6(3 p 3 2 10) 0 log 6(14 2 5 p 3)log 6(21) 0 log 6(21)Because log 6(21) is not defined, 3 is not a solution.So, there is no solution.Algebra 2Worked-Out Solution Key441
Chapter 7, continued31. log 8(5 2 12x) 5 log 8(6x 2 1)5 2 12x 5 6x 2 16 5 18x1}3 5 xCheck: log 8(5 2 12x) 5 log 8(6x 2 1)32. log 4x 5 21log 8 15 2 12 p }1 032 log8 16 p }1 3 2 1 24 log 4 x 5 4 21x 5 1 }4Check: log 4x 5 log 41 }4log 81 5 log 81 ✓Because 4 21 5 }1 4 , log 1}4 4 5 21 ✓33. 5 ln x 5 35ln x 5 7e ln x 5 e 7x 5 e 7 ø 1096.63Check: 5 ln x 5 5 ln e 7 5 5 p 7 5 35 ✓34. 1 }3 log 5 12x 5 2log 512x 5 65 log 5 12x 5 5 612x 5 15,625x 5 15,625 }12Check: 1 }3 log 5 12x 5 21}3 log 5 12 1 }15,62512 2 0 21}3 log 5 15,625 0 2log 5(15,625) 1/3 0 235. 5.2 log 42x 5 16log 525 5 2 ✓log 42x ø 3.07694 log 4 2x ø 4 3.07692x ø 71.20x ø 35.60Check: 5.2 log 42x 5 165.2 log 42 p 35.60 0 165.2log 471.2 0 16log 71.25.2 1} 0log 4 2 165.2(3.0769) 0 1616.0 0 16 ✓36. log 2(x 2 4) 5 62 log 2 (x 2 4) 5 2 6x 2 4 5 64x 5 68Check: log 2(x 2 4) 5 log 2(68 2 4) 5 log 264Because 2 6 5 64, log 264 5 6.37. log 2x 1 log 2(x 2 2) 5 3log 2[x(x 2 2)] 5 32 log 2 [x(x 2 2)] 5 2 3x(x 2 2) 5 8x 2 2 2x 2 8 5 0(x 2 4)(x 1 2) 5 0x 5 4 or x 5 22Check: log 2x 1 log 2(x 2 2) 5 3log 24 1 log 2(4 2 2) 0 3log 24 1 log 22 0 3log 28 0 3log 22 3 0 33 5 3 ✓log 2x 1 log 2(x 2 2) 5 3log 2(22) 1 log 2(22 2 2) 0 3log 2(22) 1 log 2(24) 0 3Because log 2(22) and log 2(24) are not defined, 22 isnot a solution.38. log 4(2x) 1 log 4(x 1 10) 5 2log 4[2x(x 1 10)] 5 24 log4 [2x(x 1 10)] 5 4 22x(x 1 10) 5 162x 2 2 10x 5 160 5 x 2 1 10x 1 160 5 (x 1 8)(x 1 2)x 5 28 or x 5 22Check: log 4(2x) 1 log 4(x 1 10) 5 2log 4(2(28)) 1 log 4(28 1 10) 0 2log 48 1 log 42 0 2log 416 0 2log 44 2 0 2log 4(2x) 1 log 4(x 1 10) 5 2log 4(2(22)) 1 log 4(22 1 10) 0 2log 42 1 log 48 0 2The solutions are 28 and 22.log 416 0 2log 44 2 0 22 5 2 ✓2 5 2 ✓Copyright © by McDougal Littell, a division of Houghton Mifflin Company.442Algebra 2Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.Chapter 7, continued39. ln (x 1 3) 1 ln x 5 1ln [x(x 1 3)] 5 1e ln [x(x 1 3)] 5 e 1x(x 1 3) 5 ex 2 1 3x 5 ex 2 1 3x 2 e 5 0x 5 2b 6 Ï} b 2 2 4ac}} 5 23 6 Ï } 9 2 4(2e)}}2a2x 5 23 2 Ï} 9 1 4e}}2Check: ln (x 1 3) 1 ln x 5 1ln 1 23 2 Ï} 9 1 4e}}2ln 1 3 2 Ï} 9 1 4e}or x 5 23 1 Ï} 9 1 4e}}21 3 2 1 ln 1 23 2 Ï} 9 1 4e}}22 0 122 1 ln 1 23 2 Ï} 9 1 4e}}5 23 6 Ï} 9 1 4e}}222 0 1ln (20.73) 1 ln (23.73) 0 1Because ln (20.73) and ln (23.73) are not defined,3 2 Ï } 9 1 4e} is not a solution.2ln 1 23 1 Ï} 9 1 4e}}2ln 1 3 1 Ï} 9 1 4e}ln (x 1 3) 1 ln x 5 11 3 2 1 ln 1 23 1 Ï} 9 1 4e}}22 0 122 1 ln 1 23 1 Ï} 9 1 4e}}So, the solution is 23 1 Ï} 9 1 4e}}.240. 4 ln (2x) 1 3 5 214 ln (2x) 5 18ln(2x) 5 9 }2e ln (2x) 5 e 9/22x ø e 9/2x 5 2e 9/2 ø 290.02Check: 4 ln(2x) 1 3 0 214 ln(2(2e 9/2 )) 1 3 0 214 ln(e 9/2 ) 1 3 0 2141 }9 22 1 3 0 2121 5 21 ✓41. log 5(x 1 4) 1 log 5(x 1 1) 5 2log 5[(x 1 4)(x 1 1)] 5 25 log 5 [(x 1 4)(x 1 1)] 5 5 2(x 1 4)(x 1 1) 5 25x 2 1 5x 1 4 5 25x 2 1 5x 2 21 5 022 0 129 1 9 1 4eln 1 }4 2 0 1ln (e) 0 11 5 1 ✓x 5 256 Ï }}25 2 4(221)}}2x 5 25 2 Ï} 109}25 256 Ï} 109}2or x 5 25 1 Ï} 109}2Check: log 5(x 1 4) 1 log 5(x 1 1) 5 2log 5 1 25 2 Ï} 109}21 42 1 log 5 1 25 2 Ï} 109}log 5 1 3 2 Ï} 109}22 1 log 5 1 23 2 Ï} 109}21 1 2 0 222 0 2log 5(27.44) 1 log 5(26.72) 0 2Because log 5(27.44) and log 5(26.72) are not defined,25 2 Ï } 109} is not a solution.2log 5 1 25 1 Ï} 109}2log 5(x 1 4) 1 log 5(x 1 1) 5 21 42 1 log 5 1 25 1 Ï} 109}log 5 1 3 1 Ï} 109}So, 25 1 Ï} 109} is a solution.242. log 63x 1 log 6(x 2 1) 5 3log 6[3x(x 2 1)] 5 36 log 6 [3x(x 2 1)] 5 6 322 1 log 5 1 23 1 Ï} 109}3x(x 2 1) 5 2163x 2 2 3x 5 2163x 2 2 3x 2 216 5 0x 2 2 x 2 72 5 0(x 2 9)(x 1 8) 5 02x 5 9 or x 5 28Check: log 63x 1 log 6(x 2 1) 5 3log 6(3 p 9) 1 log 6(9 2 1) 0 3log 627 1 log 68 0 3log 6216 0 3log 66 3 0 33 5 3 ✓log 63x 1 log 6(x 2 1) 5 3log 6(3(28)) 1 log 6(28 2 1) 0 3log 6(224) 1 log 6(29) 0 31 1 2 0 222 0 2log 5 1 109 2 9 }4 2 0 2log 525 0 22 5 2 ✓Because log 6(224) and log 6(29) are not defined, 28 isnot a solution.The solution is 9.Algebra 2Worked-Out Solution Key443
Chapter 7, continued43. log 3(x 2 9) 1 log 3(x 2 3) 5 2log 3[(x 2 9)(x 2 3)] 5 23 log 3 [(x 2 9)(x 2 3)] 5 3 2(x 2 9)(x 2 3) 5 9x 2 2 12x 1 27 5 9x 2 2 12x 1 18 5 0x 5 12 6 Ï }}144 2 4(18)}} 5 126 Ï} 72} 5 663 Ï } 222x 5 6 2 3 Ï } 2 or x 5 6 1 3 Ï } 2Check: log 3(x 2 9) 1 log 3(x 2 3) 5 2log 3(6 2 3 Ï } 2 2 9) 1 log 3(6 2 3 Ï } 2 2 3) 0 2log 3(23 2 3 Ï } 2 ) 1 log 3(3 2 3 Ï } 2 ) 0 2log 3(27.25) 1 log 3(21.24) 0 2Because log 3(27.25) and log 3(21.24) are notdefined, 6 2 3 Ï } 2 is not a solution.log 3(x 2 9) 1 log 3(x 2 3) 5 2log 3(6 1 3 Ï } 2 2 9) 1 log 3(6 1 3 Ï } 2 2 3) 0 2log 3[(3 Ï } 2 ) 2 2 32] 1 log 3(3 1 3 Ï } 2 ) 0 2So, the solution is 6 1 3 Ï } 2 .44. A; 3log 8(2x 1 7) 1 8 5 103log 8(2x 1 7) 5 2log 8(2x 1 7) 5 2 }38 log 8 (2x 1 7) 5 8 2/32x 1 7 5 4log 3[(3 Ï } 2 ) 2 2 3 2 ] 0 22x 5 23x 5 2 3 }2log 33 2 0 2or 21.52 5 2 ✓45. The error was made in simplifying xlog 36 as 2x;log 36 ø 1.631, not 2.3 x 1 1 5 6xlog 33 x 1 1 5 log 36 xx 1 1 5 log 36xx 1 1 5 xlog 36x 1 1 ø 1.631x1 ø 0.631x1.58 ø x46. When exponentiating both sides, base 3 should have beenused, not base e.log 310x 5 53 log 3 10x 5 3 510x 5 243x 5 24.347. Sample answer: 2 x 5 16 and log 8(5 1 x) 5 1 }348. 3 x 1 4 5 6 2x 2 5log 33 x 1 4 5 log 36 2x 2 5x 1 4 5 (2x 2 5)(log 36)x 1 4 5 2x log 36 2 5 log 364 1 5 log 36 5 2x log 36 2 x4 1 5 log 36 5 x(2 log 36 2 1)4 1 5 log 36}2 log 36 2 1 5 x4 1 5 p }log 6log 3}2 p }log 6 5 xlog 3 2 14 1 5(1.631)}2(1.631) 2 1 ø x5.37 ø x49. 10 3x 2 8 5 2 5 2 xlog10 3x 2 8 5 log2 5 2 x3x 2 8 5 (5 2 x) log 23x 2 8 5 5 log 2 2 x log 2x(3 1 log 2) 5 8 1 5 log 28 1 log 32x 5 }3 1 log 2x ø 2.8850. log 2(x 1 1) 5 log 83xlog 2(x 1 1) 5 log 2 (3x)}log 28log 2(x 1 1) 5 log 2 (3x)}33log 2(x 1 1) 5 log 2(3x)log 2(x 1 1) 3 5 log 2(3x)(x 1 1) 3 5 3xx 3 1 3x 2 1 3x 1 1 5 3xx 3 1 3x 2 1 1 5 0Graphing the function y 5 x 3 1 3x 2 1 1 shows that thex-intercept is negative. Because the log of a negativenumber is not defined, there is no solution.51. log 3x 5 log 96xlog 3x 5 log 3 6x}log 39log 3x 5 log 3 6x}22 log 3x 5 log 36xlog 3x 2 5 log 36xx 2 5 6xx(x 2 6) 5 0x 5 0 or x 5 6Check: log 3x 5 log 96xlog 30 0 log 96(0)log 30 0 log 90Because log 30 and log 90 are not defined, 0 is not asolution.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.444Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.log 3x 5 log 96xlog 36 0 log 9(6 p 6)log 36 0 log 936log 6 log 36} 0 }log 3 log 9log 6}log 3 5 }log 6log 3 ✓5log 62}log 3 2So, 6 is the solution.52. 2 2x 2 12 p 2 x 1 32 5 0(2 x 2 4)(2 x 2 8) 5 02 x 5 4 or 2 x 5 8x 5 log 24 or x 5 log 28x 5 log 22 2 or x 5 log 22 3x 5 2 or x 5 3Check: 2 2x 2 12 p 2 x 1 32 5 02 2(2) 2 12 p 2 2 1 32 0 016 2 48 1 32 0 02 2x 2 12 p 2 x 1 32 5 02 2(3) 2 12 p 2 3 1 32 0 064 2 96 1 32 0 00 5 0 ✓So, the solutions are 2 and 3.53. 5 2x 1 20 p 5 x 2 125 5 0(5 x 1 25)(5 x 2 5) 5 00 5 0 ✓5 x 5 225 or 5 x 5 5x 5 log 5(225) x 5 log 55Not defined x 5 1Check: 5 2x 1 20 p 5 x 2 125 5 05 2(1) 1 20 p 5 1 2 125 0 0So, 1 is the solution.Problem Solving25 1 100 2 125 0 00 5 0 ✓54. T 5 100, T 05 200, T R5 75, and r 5 0.054T 5 (T 02 T R)e 2rt 1 T R100 5 (200 2 75)e 20.054t 1 7525 5 125e 20.054t0.2 5 e 20.054tln 0.2 5 ln e 20.054t21.6094 ø 20.054t30 ø tIt will take about 30 minutes to cool the beef stew.55. T 5 37, T 05 75, t 5 1, and r 5 1.37T 5 (T 02 T R)e 2rt 1 T R37 5 (75 2 T R)e 21.37(1) 1 T R37 5 75e 21.37 2 T Re 21.37 1 T R37 2 75e 21.37 5 T R(2e 21.37 1 1)37 2 75e}}21.372e 21.37 1 1 5 T R37 2 75(0.25)}}20.25 1 1 ø T R24 ø T RThe outdoor temperature is about 248F.56. A 5 P 11 1 }n2 r nt When P 5 100, r 5 0.06, and A 5 1000a. When n 5 1: 1000 5 10011 1 }0.061 2 (1)t10 5 1.06 tlog10 5 log1.06 t1}log1.06 5 t1 5 tlog1.0639.5 ø tIt will take about 39.5 years for the balance toreach $1000.b. When n 5 4: 1000 5 10011 1 }0.064 2 (4)t10 5 1.015 4tlog 10 5 log 1.015 4t1}4 log 1.015 5 t1 5 4t log 1.01538.7 ø tIt will take about 38.7 years for the balance toreach $1000.c. When n 5 365: 1000 5 10011 1 }0.06365 2 (365)t10 5 1 }365.06365 2 365tlog 10 5 log 1 }365.06365 2 365t1}}365log 1 }365.06 5 t365 21 5 365t log 1 365.06 }365 238.4 ø tIt will take about 38.4 years for the balance toreach $1000.Algebra 2Worked-Out Solution Key445
Chapter 7, continued57. When R 5 5: R 5 100e 20.00043t5 5 100e 20.00043t0.05 5 e 20.00043tln 0.05 5 ln e 20.00043t22.99573 ø 20.00043t6967 ø tIt will take about 6967 years for only 5 grams of radiumto be present.58. C; A 5 3(800) 5 2400, r 5 0.0225, and P 5 800A 5 Pe rt2400 5 800e 0.0225t3 5 e 0.0225tln 3 5 ln e 0.0225t1.099 ø 0.0225t48.8 ø t59. a. By changing the window settings and using thetrace feature, you can estimate the amount of energyreleased by each earthquake:Ocotillo Wells, CA: about 24,000 kilowatt hoursAthens: about 11,500,000 kilowatt hoursFukuoka: about 127,000,000 kilowatt hours.b. Ocotillo Wells, CA: R 5 0.67 log (0.37E) 1 1.464.1 5 0.67 log (0.37E) 1 1.462.64 5 0.67 log (0.37E)2.64} 5 log (0.37E)0.6710 (2.64/0.67) log (0.37E)5 1010 (2.64/0.67) 5 0.37E8715.62 ø 0.37E23,555.7 ø EThe amount of energy released at Ocotillo Wells, CA,was about 23,556 kilowatt-hours.Athens: R 5 0.67 log (0.37E) 1 1.465.9 5 0.67 log (0.37E) 1 1.464.44 5 0.67 log (0.37E)4.44} 5 log (0.37E)0.6710 (4.44/0.67) log (0.37E)5 1010 (4.44/0.67) 5 0.37E4235119.52 ø 0.37E11,446,269.0 ø EThe amount of energy released at Athens was about11,446,269 kilowatt-hours.Fukuoka: R 5 0.67 log (0.37E) 1 1.466.6 5 0.67 log (0.37E) 1 1.465.14 5 0.67 log (0.37E)5.14} 5 log (0.37E)0.6710 (5.14/0.67) log (0.37E)5 1010 (5.14/0.67) 5 0.37E46,950,669.66 ø 0.37E126,893,701.8 ø EThe amount of energy released at Fukuoka was about126,893,702 kilowatt-hours.60. a. I(x) 5 0.3I 0; m 5 0.43I(x) 5 I 0e 2mx0.3I 05 I 0e 20.43x0.3 5 e 20.43xln 0.3 5 ln e 20.43xln 0.3 5 20.43x21.20 ø 20.43x2.8 ø xThe thickness should be about 2.8 centimeters.b. I(x) 5 0.3I 0; m 5 3.2I(x) 5 I 0e 2mx0.3I 05 I 0e 23.2x0.3 5 e 23.2xln 0.3 5 ln e 23.2xln 0.3 5 23.2x21.20 ø 22.3x0.375 ø xThe thickness should be about 0.38 centimeters.c. I(x) 5 0.3I 0; m 5 43I(x) 5 I 0e 2mx0.3I 05 I 0e 243x0.3 5 e 243xln 0.3 5 ln e 243xln 0.3 5 243x21.20 ø 243x0.03 ø xThe thickness should be about 0.03 centimeters.d. Lead is a better material to use than aluminum orcopper because it can be much thinner and still shieldthe same amount of X-rays.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.446Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.25661. When h 5 200: h(t) 5 }1 1 13e 20.65t256200 5 }1 1 13e 20.65t1 1 13e 20.65t 5 256 }20013e 20.65t 5 0.28e 20.65t 5 0.28 }13ln e 20.65t ø ln 1 0.28 }13 220.65t 5 ln 1 0.28 }13 220.65t 5 23.84t ø 5.9It takes about 6 weeks for the seedling to reach a heightof 200 centimeters.62. 3x 2 y 5 7 3x 2 y 5 7x 1 2y 5 14 3 23 23x 2 6y 5 2423x 2 5 5 7 → x 5 4The solution is (4, 5).27y 5 235y 5 563. 5x 2 y 5 7 3 5 25x 2 5y 5 352x 1 5y 5 23 2x 1 5y 5 2351 }32 272 2 y 5 7 → y 5 2 }29 27The solution is 1 }32 27 , 2 }29 272 .27x 5 32x 5 32 }2764. x 1 4y 5 26 3 2 2x 1 8y 5 21222x 1 y 5 12 22x 1 y 5 12x 1 4(0) 5 26 → x 5 26The solution is (26, 0).9y 5 0y 5 065. f(x) 5 x 3 2 2x 2 1 5; 2 or 0 positive real zerosf(2x) 5 2x 3 2 2x 2 1 5; 1 negative real zeroPositivereal zerosNegativereal zerosImaginaryzeros2 1 00 1 266. f(x) 5 x 4 1 6x 3 2 x 2 1 7x 2 8;3 or 1 positive real zerosf(2x) 5 x 4 2 6x 3 2 x 2 2 7x 2 8; 1 negative real zeroPositivereal zerosNegativereal zerosImaginaryzeros3 1 01 1 267. f(x) 5 x 5 2 3x 3 1 7x 2 1 6x 1 9;2 or 0 positve real zerosf(2x) 5 2x 5 1 x 3 1 7x 2 2 6x 1 9;3 or 1 negative real zerosPositivereal zerosNegativereal zerosImaginaryzeros2 3 02 1 20 3 20 1 468. f(x) 5 x 7 1 10x 6 2 5x 4 1 12x 3 2 17;3 or 1 positive real zerosf(2x) 5 2x 7 1 10x 6 2 5x 4 2 12x 3 2 17;2 or 0 negative real zerosPositivereal zerosNegativereal zerosImaginaryzeros3 2 23 0 41 2 41 0 669. f(1) f(2) f(3) f(4) f(5) f(6)19 28 27 16 25 2369 21 211 221 231210 210 210 210f(x) 5 ax 2 1 bx 1 c(1, 19): a 1 b 1 c 5 19(2, 28): 4a 1 2b 1 c 5 28(3, 27): 9a 1 3b 1 c 5 27Using a calculator, the solution is a 5 25, b 5 24,and c 5 0. So, a function that fits the data isf(x) 5 25x 2 1 24x.Algebra 2Worked-Out Solution Key447
Chapter 7, continued70. f(1) f(2) f(3) f(4) f(5) f(6)0 2 12 36 80 1502 10 24 44 708 14 20 266 6 6f(x) 5 ax 3 1 bx 2 1 cx 1 d(1, 0): a 1 b 1 c 1 d 5 0(2, 2): 8a 1 4b 1 2c 1 d 5 2(3, 12): 27a 1 9b 1 3c 1 d 5 12(4, 36): 64a 1 16b 1 4c 1 d 5 36Using a calculator, the solution is a 5 1, b 5 22,c 5 1, and d 5 0. So a function that fits the datais f(x) 5 x 3 2 2x 2 1 x.Problem Solving Workshop 7.6 (p. 525)1. 8 2 2e 3x 5 2145. log 25x 5 2X Y1.4 1.5 1.3219.6 1.585.7 1.8074.8 2.9 2.16991 2.3219X=.8The solution is 0.8.6. log (23x 1 7) 5 1X Y1-1.2 1.0253-1.1 1.0128-1 1-.9 .98677-.8 .97313-.7 .95904-.6 .94448X=-1The solution is 21.7. 4 ln x 1 6 5 12IntersectionX=.8 Y=2IntersectionX=-1 Y=1X Y1.4 1.3598.5 -.9634.6 -4.099.7 -8.332.8 -14.05.9 -21.761 -32.17X=.8IntersectionX=.79929842 Y=-14X Y14.1 11.6444.2 11.744.3 11.8344.4 11.9264.5 12.0164.6 12.1044.7 12.19X=4.5IntersectionX=4.4816891 Y=12The solution is about 0.8.2. 7 2 10 5 2 x 5 29The solution is about 4.48.8. 11 log (x 1 9) 2 5 5 8X Y13.5 -24.623.6 -18.123.7 -12.953.8 -8.8493.9 -5.5894 -34.1 -.9433X=3.8The solution is about 3.8.3. e 5x 2 8 1 3 5 15X Y11.8 5.71831.9 7.48172 10.3892.1 15.1822.2 23.0862.3 36.1152.4 57.598X=2.1The solution is about 2.1.4. 1.6(3) 24x 1 5.6 5 6X Y10 7.2.1 6.631.2 6.2644.3 6.0281.4 5.8759.5 5.7778.6 5.7146X=.3The solution is about 0.315.IntersectionX=3.79588 Y=-9IntersectionX=2.0969813 Y=15IntersectionX=.31546788 Y=6X Y15.8 7.87295.9 7.9056 7.9376.1 7.96876.2 8.00036.3 8.03166.4 8.0627X=6.2The solution is about 6.2.9. y 5 8882(1.04) xX Y12.7 9874.22.8 99132.9 9951.93 99913.1 100303.2 100703.3 10109X=3IntersectionX=6.199108 Y=8IntersectionX=3.0228501 Y=10000The gross national product was $10 trillion in2001 (x ø 3).10. You could make the step between values even smaller tofind the solution of 4 x 5 11 more precisely.11. y 5 5 log x 1 2X Y1240 13.901245 13.946250 13.99255 14.033260 14.075265 14.116270 14.157X=250IntersectionX=251.18864 Y=14Copyright © by McDougal Littell, a division of Houghton Mifflin Company.To reveal stars of magnitude 14, the diameter ofthe telescope’s objective lense must be about251 millimeters.448Algebra 2Worked-Out Solution Key
Chapter 7, continued12. A 5 P 1 1 1 r } n 2 nt , A 5 6000, P 5 5000, r 5 0.03, andn 5 4y 5 500011 1 }0.034 2 4xX Y15.8 5946.45.9 5964.26 5982.16.1 60006.2 6017.96.3 6035.96.4 6054X=6.1IntersectionX=6.100147 Y=60004. 263(0.96) x < 227X Y120.73 -27.0320.74 -27.0220.75 -27.0120.76 -2720.77 -26.9820.78 -26.9720.79 -26.96X=20.76The solution is x < 20.765. 95(1.6) x ≤ 1620IntersectionX=20.755915Y=-27The balance will reach $6000 after about 6.1 years.0.1226x 2 7.82813. y 5 1.0245 2 eX Y1-2 1.0242-1 1.02410 1.02411 1.0242 1.0243 1.02394 1.0238X=0IntersectionX=.03225112 Y=1.0241Deep water in the South Atlantic Ocean off Antarcticahas a temperature of about 0.038C.7.6 Extension (p. 527)1. 3 x ≤ 20X Y12.68 18.9972.69 19.2072.7 19.4192.71 19.6342.72 19.852.73 20.072.74 20.291X=2.72IntersectionX=2.726833 Y=20X Y16 1593.86.01 1601.36.02 1608.96.03 1616.56.04 1624.16.05 1631.76.06 1639.4X=6.03The solution is x ≤ 6.03.6. 22841 }9 72 x > 2135X Y1-2.98 -134.3-2.97 -134.6-2.96 -135-2.95 -135.3-2.94 -135.7-2.93 -136-2.92 -136.3X=-2.96The solution is x < 22.96.7. log 3x ≥ 3IntersectionX=6.0346439 Y=1620IntersectionX=-2.959239Y=-135Copyright © by McDougal Littell, a division of Houghton Mifflin Company.The solution is x ≤ 2.727.2. 281 }2 32 x > 9X Y12.75 9.18142.76 9.14422.77 9.10722.78 9.07042.79 9.03372.8 8.99712.81 8.9607X=2.79The solution is x < 2.799.3. 244(0.35) x ≥ 50X Y11.48 51.5951.49 51.0571.5 50.5231.51 49.9961.52 49.4741.53 48.9571.54 48.446X=1.51The solution is x ≤ 1.51.IntersectionX=2.7992049 Y=9IntersectionX=1.5099179 Y=50X Y125 2.929926 2.965627 - 328 3.033129 3.06530 3.095931 3.1257X=27The solution is x ≥ 27.8. log 5x < 2X Y121 1.891722 1.920623 1.948224 1.974625 226 2.024427 2.0478X=25The solution is 0 < x < 25.9. log 6x 1 9 ≤ 11X Y132 10.93433 10.95134 10.96835 10.98436 1137 11.01538 11.03X=36IntersectionX=27 Y=3IntersectionX=25 Y=2IntersectionX=36 Y=11The solution is 0 < x ≤ 36.Algebra 2Worked-Out Solution Key449
Chapter 7, continued10. 2 log 4x 2 1 > 4X Y130 3.906931 3.954232 433 4.044434 4.087535 4.129336 4.1699X=32The solution is x > 32.11. 24 log 2x > 220X Y128 -19.2329 -19.4330 -19.6331 -19.8232 -2033 -20.1834 -20.35X=32The solution is 0 < x < 32.12. 0 ≤ log 7x ≤ 1X Y11 02 .356213 .564584 .712415 .827096 .920787 1X=1The solution is 1 ≤ x ≤ 7.IntersectionX=32 Y=4IntersectionX=32 Y=-20IntersectionX=7 Y=113. A 5 P 1 1 1 r } n 2 nt ; P 5 1000; r 5 0.035; n 5 12A 5 100011 1 }0.03512 2 12tX Y15 1190.95.05 11935.1 1195.15.15 1197.25.2 1199.35.25 1201.45.3 1203.5X=5.25IntersectionX=5.2167804 Y=1200Your balance is at least $1200 after about 5.25 years.ln 214. t 5 }ln (1 1 r)X Y1.069 10.388.07 10.245.071 10.105.072 9.9696.073 9.8377.074 9.7093.075 9.5844X=.072IntersectionX=.07177346 Y=10The values of t are less than 10 for values of r greaterthan about 0.072. So, rates of return r > 7.2% will doublethe value of an investment in less than 10 years.Lesson 7.7Investigating Algebra Activity 7.7 (p. 528)ExploreAnswers will vary.Draw conclusions1. The initial number of pennies was 100.You would expect the number of penniesremaining to decrease by 50% after eachtoss.2. An equation that should model the datais y 5 100(0.5) x .3. Answers will vary.4. Answers will vary.7.7 Guided Practice (pp. 531 – 533)1. (1, 6): 6 5 ab 1 → a 5 }6 b(3, 24): 24 5 ab 324 5 1 }6 b2 b 324 5 6b 24 5 b 22 5 ba 5 }6 b 5 }6 2 5 3So, an equation is y 5 3 p 2 x .2. (2, 8): 8 5 ab 2 → a 5 }8 b 2(3, 32): 32 5 ab 332 5 1 8 }b 2 2 b332 5 8b4 5 ba 5 }8 b 5 8 2 }4 5 8 2 }16 5 }1 2So, an equation is y 5 1 }2 p 4x .3. (3, 8): 8 5 ab 3 → a 5 }8 b 3(6, 64): 64 5 ab 664 5 1 8 }b 3 2 b664 5 8b 38 5 b 32 5 ba 5 }8 b 5 8 3 }2 5 8 3 }8 5 1So, an equation is y 5 2 x .Copyright © by McDougal Littell, a division of Houghton Mifflin Company.450Algebra 2Worked-Out Solution Key
Chapter 7, continued4. x 1 2 3 4 5 6 7lny 2.71 3.14 3.69 3.95 4.38 4.65 4.94The line appears to pass through (1, 2.80) and(6, 4.65).4.65 2 2.80M 5 } ø 0.376 2 1ln yln y 2 2.80 5 0.37(x 2 1)ln y 5 0.37x 1 2.430.37x 1 2.43y 5 ey 5 e 2.43 (e 0.37 ) xy 5 11.36(1.45) x1(1, 2.80)1(6, 4.65)Using a scatter plot of (x, ln y), an exponential model forthe data (x, y) is y 5 11.36(1.45) x . Using the exponentialregression feature of a graphing calculator, a model isy 5 11.39(1.45) x .x7. (5, 8): 8 5 a p 5 b → a 5 }85 b(10, 34): 34 5 a p 10 b34 5 1 8 }5 b 2 p 10b34 5 8 p 2 b17}4 5 2blog 217}4 5 blog }174}log 2 5 b2.09 ø ba 5 }85 ø 8b } ø 0.2772.095So, an equation is y 5 0.277x 2.09 .Copyright © by McDougal Littell, a division of Houghton Mifflin Company.5. (2, 1): 1 5 a p 2 b → a 5 }12 b(7, 6): 6 5 a p 7 b6 5 1 1 }2 b 2 p 7b6 5 1 p 1 }7 22log 7/26 5 blog 65 b}log 7 }21.43 ø ba 5 }12 ø 1b } ø 0.3711.432So, an equation is y 5 0.371x 1.43 .6. (3, 4): 4 5 a p 3 b → a 5 }43 b(6, 15): 15 5 a p 6 bb15 5 1 4 }3 b 2 p 6b15 5 4 p 2 b15}4 5 2blog 215}4 5 blog }154}log 2 5 b1.91 ø ba 5 }43 ø 4b } ø 0.4911.913So, an equation is y 5 0.491x 1.91 .8. (3, 5): 5 5 a p 3 b → a 5 }53 b(3, 7): 7 5 a p 3 b7 5 1 5 }3 b 2 p 3b7 Þ 5 p 1Sample answer: The two points cannot form a powerfunction.9.ln x 1.099 2.398 2.944 3.612 4.007ln y 5.196 4.583 4.154 3.661 3.3501ln y(1.099, 5.196)1(4.007, 3.350)ln x3.350 2 5.196M 5 }}4.007 2 1.099 ø 20.635ln y 2 3.350 5 20.635(ln x 2 4.007)ln y 5 20.635 ln x 1 5.894ln y 5 ln x 20.635 1 5.894y 5 e ln x20.635 1 5.894y 5 e 5.894 p e lnx20.635y 5 362.85x 20.6357.7 Exercises (pp. 533 – 536)Skill Practice1. Given a set of more than two data pairs (x, y), you candecide whether an exponential function fits the data wellby making a scatterplot of the points (x, ln y).Algebra 2Worked-Out Solution Key451
Chapter 7, continued2. You can determine whether a power function is a goodmodel for a set of data pairs (x, y) by determiningwhether the set of transformed points (ln x, ln y) fits alinear pattern.3. (1, 3): 3 5 a p b 1 → a 5 }3 b(2, 12): 12 5 a p b 212 5 1 }3 b2 p b 212 5 3b4 5 ba 5 3 }4 5 0.75So, an equation is y 5 3 }4 p 4x .4. (2, 24): 24 5 a p b 2 → a 5 24 }b 2(3, 144): 144 5 a p b 3144 5 1 24 }b 2 2 b3144 5 24b6 5 ba 5 }24 b 5 242 }36 5 }2 3So, an equation is y 5 2 }3 p 6x .5. (3, 1): 1 5 a p b 3 → a 5 1 }b 3(5, 4): 4 5 a p b 54 5 1 1 }b 3 2 p b54 5 b 22 5 ba 5 }1 b 5 1 3 }2 5 1 3 }8So, an equation is y 5 1 }8 p 2x .6. (3, 27): 27 5 a p b 3 → a 5 27 }b 3(5, 243): 243 5 a p b 5243 5 1 27 }b 3 2 p b5243 5 27b 29 5 b 23 5 ba 5 }27 b 5 273 }3 5 1 3So, an equation is y 5 1 p 3 x 5 3 x .7. (1, 2): 2 5 a p b 1 → a 5 2 }b(3, 50): 50 5 a p b 350 5 1 }2 b2 b 350 5 2b 225 5 b 25 5 ba 5 }2 b 5 }2 5So, an equation is y 5 2 }5 p 5x .8. (1, 40): 40 5 a p b 1 → a 5 40 }b(3, 640): 640 5 a p b 3640 5 1 40 }b 2 b 3640 5 40b 216 5 b 24 5 ba 5 40 }b 5 40 }4 5 10So, an equation is y 5 10 p 4 x .9. (21, 10): 10 5 a p b 21 → a 5 10 }b 21(4, 0.31): 0.31 5 a p b 40.31 5 1 10 }b 21 2 b40.31 5 10b 50.031 5 b 55Ï } 0.031 5 b0.5 ø ba 5 }10b 5 1021 }So an equation is y 5 5 p 1 }1 22 x .( 5 Ï } 0.031 ) 21 ø 510. (2, 6.4): 6.4 5 a p b 2 → a 5 6.4 }b 2(5, 409.6): 409.6 5 a p b 5409.6 5 1 6.4 }b 2 2 b5409.6 5 6.4b 364 5 b 34 5 ba 5 }6.4b 5 6.42 }4 5 0.4 2So an equation is y 5 0.4 p 4 x .Copyright © by McDougal Littell, a division of Houghton Mifflin Company.452Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.11. x 1 2 3 4 512.13.14.ln y 2.89 3.58 4.28 4.97 5.66ln y(4, 4.97)(2, 3.58)121x4.97 2 3.58M 5 } ø 0.704 2 2ln y 5 0.70x 1 2.180.70x 1 2.18y 5 ey 5 e 2.18 (e 0.7 ) xy 5 8.85(2.01) xx 1 2 3 4 5ln y 1.19 2.31 3.42 4.53 5.64ln y(4, 4.53)(2, 2.31)121x4.53 2 2.31M 5 } 5 1.114 2 2ln y 5 1.11x 1 0.091.11x 1 0.09y 5 ey 5 e 0.09 (e 1.11 ) xy 5 1.09(3.03) xx 1 2 3 4 5ln y 2.28 2.50 2.72 2.94 3.17ln y(5, 3.17)(2, 2.50)121x3.17 2 2.50M 5 } ø 0.225 2 2ln y 5 0.22x 1 2.060.22x 1 2.06y 5 ey 5 e 2.06 (e 0.22 ) xln y 2 3.58 5 0.70(x 2 2)ln y 2 2.31 5 1.11(x 2 2)ln y 2 2.50 5 0.22(x 2 2)x 1 2 3 4 5ln y 0.34 1.90 3.49 5.08 6.67211ln y(1, 0.34)(4, 5.08)5.08 2 0.34M 5 } 5 1.584 2 1ln y 2 0.34 5 1.58(x 2 1)ln y 5 1.58x 2 1.241.58x 2 1.24y 5 ey 5 e 21.24 (e 1.58 ) xy 5 0.289(4.85) x15. (4, 3): 3 5 a p 4 b → a 5 3 }4 b(8, 15): 15 5 a p 8 b15 5 1 3 }4 b 2 p 8b15 5 3 p 2 b5 5 2 blog 25 5 blog 5}log 2 5 b2.32 ø ba 5 }34 ø 3b } ø 0.122.324So, an equation is y 5 0.12x 2.32 .16. (5, 9): 9 5 a p 5 b → a 5 9 }5 b(8, 34): 34 5 a p 8 b34 5 1 9 }5 b 2 8b34 5 9 p 1 }8 52 b34}9 5 1 }8 52 blog 8/534}9 5 blog }34 9}log }8 55 b2.83 ø ba 5 }95 ø 9b } ø 0.092.835So, an equation is y 5 0.09x 2.83 .xAlgebra 2Worked-Out Solution Key453
Chapter 7, continued17. (2, 3): 3 5 a p 2 b → a 5 }32 b(6, 12): 12 5 a p 6 b20. (5, 10): 10 5 a p 5 b → a 5 }105 b(12, 81): 81 5 a p 12 b12 5 1 3 }2 b 2 6b12 5 3 p 3 b4 5 3 blog 34 5 blog 4}log 3 5 b1.26 ø ba 5 }32 ø 3b } ø 1.251.262So, an equation is y 5 1.25 p x 1.26 .18. (3, 14): 14 5 a p 3 b → a 5 }143 b(9, 44): 44 5 a p 9 b44 5 1 14 }3 b 2 9b44 5 14 p 3 b44}14 5 3blog 3 44 }14 5 blog }4414}log 3 5 b1.04 ø ba 5 }143 ø 14 b } ø 4.471.043So, an equation is y 5 4.47x 1.04 .19. (4, 8): 8 5 a p 4 b → a 5 }84 b(8, 30): 30 5 a p 8 b30 5 1 8 }4 b 2 p 8b30 5 8 p 2 b15}4 5 2blog 215}4 5 blog }154}log 2 5 b1.91 ø ba 5 }84 ø 8b } ø 0.571.914So, an equation is y 5 0.57x 1.91 .81 5 1 10 }5 b 2 12b81 5 10 p 1 }12 5 2 b8.1 5 (2.4) blog 248.1 5 blog 8.1}log 2.4 5 b2.39 ø ba 5 }105 ø 10 b } ø 0.212.395So, an equation is y 5 0.21x 2.39 .21. (4, 6.2): 6.2 5 a p 4 b → a 5 }6.24 b(7, 23): 23 5 a p 7 b23 5 1 6.2 }4 b 2 7b23 5 6.21 }7 42 b23}6.2 5 1 }7 42 blog 7/423}6.2 5 blog }236.2}log }7 5 b42.34 ø ba 5 }6.24 ø 6.2b } ø 0.242.344So, an equation is y 5 0.24x 2.34 .22. (3.1, 5): 5 5 a p 3.1 b 5→ a 5 }3.1 b(6.8, 9.7): 9.7 5 a p 6.8 b9.7 5 1 5 }3.1 b 2 6.8b9.7 5 51 }6.83.12 b1.94 5 1 }6.83.12 blog1}6.83.12 1.94 5 blog 1.94}log }6.8 5 b3.10.84 ø b5a 5 }3.1 ø 5b } ø 1.930.843.1So, an equation is y 5 1.93x 0.84 .Copyright © by McDougal Littell, a division of Houghton Mifflin Company.454Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.23.24.ln x 0 0.693 1.099 1.386 1.609ln y 20.511 1.411 2.518 3.296 3.90225.ln y(1.609, 3.902)11 ln x(0, 20.511)3.902 2 (20.511)M 5 }} ø 2.741.609 2 0ln y 5 2.74 ln x 2 0.507ln y 5 ln x 2.74 2 0.507y 5 e ln x 2.74 2 0.507y 5 e 20.507 p e ln x 2.74y 5 0.6023x 2.7426.ln y(1.386, 2.734)(0.693, 1.569)11ln x2.734 2 1.569M 5 }}1.386 2 0.693 ø 1.68ln y 5 1.68 ln x 1 0.406ln y 5 ln x 1.68 1 0.406y 5 e ln x1.68 1 0.406y 5 e 0.406 ln x 1.68p ey 5 1.5008x 1.68ln y 2 3.902 5 2.74(ln x 2 1.609)ln x 0 0.693 1.099 1.386 1.609ln y 0.406 1.569 2.251 2.734 3.105ln y 2 2.734 5 1.68(ln x 2 1.386)ln x 0 0.693 1.099 1.386 1.609ln y 0.916 1.308 1.548 1.705 1.8252ln y(0, 0.916)(1.609, 1.825)1 ln x1.825 2 0.916M 5 }} ø 0.561.609 2 0ln y 2 1.825 5 0.56(ln x 2 1.609)ln y 5 0.56 ln x 1 0.924ln y 5 ln x 0.56 1 0.924y 5 e ln x0.56 1 0.924y 5 e 0.924 ln x 0.56p ey 5 2.5193x 0.56ln x 0 0.693 1.099 1.386 1.609ln y 20.211 20.010 0.104 0.191 0.2550.1ln y(1.386, 0.191)2 ln x(0.693, 20.010)0.191 2 (20.010)M 5 }} ø 0.291.386 2 0.693ln y 2 0.191 5 0.29(ln x 2 1.386)27. A; log y 5 2x 1 1ln y 5 0.29 ln x 2 0.2109ln y 5 ln x 0.29 2 0.210910 log y 5 10 2x 1 1y 5 e ln x0.29 2 0.2109y 5 e 20.2109 ln x 0.29p ey 5 0.8099x 0.29y 5 10 2x 1 1y 5 10 1 p 10 2xy 5 10 p (10 2 ) xy 5 10(100) xAlgebra 2Worked-Out Solution Key455
Chapter 7, continued28. The expression e 2x 1 1 should have been rewritten as theproduct of e 2x and e 1 , not the sum.ln y 5 2x 1 1y 5 e 2x 1 1y 5 e 2x p e 1y 5 (e 2 ) x p ey 5 (7.39) x p 2.7229. The expression 3 ln x is equivalent to ln x 3 , not ln 3x.ln y 5 3 ln x 2 2ln y 5 ln x 3 2 2ln x3 2 2y 5 ey 5 e 22 p e lnx3y 5 0.135x 330. y 5 ab x y 5 ax bln y 5 ln ab xln y 5 ln ax bln y 5 ln a 1 x ln bln y 5 ln a 1 b ln xThe slope is ln b and The slope is b and thethe y-intercept is ln a y-intercept is ln a.Problem Solving31. a. ln C 5.153 4.905 4.538 2.565 3.332ln W 6.565 6.105 4.963 20.919 2.2732.ln x 20.949 20.324 0 0.421 1.649 2.255ln y 21.423 20.486 0 0.632 2.473 3.3831ln y21 (0, 0)(1.649, 2.473)ln xM 5 }2.4731.649 ø 1.5ln y 2 2.473 5 1.5(ln x 2 1.649)ln y 5 1.5ln x 2 0.0005ln y 5 ln x 1.5 2 0.0005y 5 e ln x1.5 2 0.0005y 5 e 20.0005 ln x1.5p ey 5 0.9995x 1.533. a.x 1 2 3 4ln y 0 0.693 1.609 1.946x 5 6 7ln y 2.996 3.689 4.382ln W(4.905, 6.105)ln y(7, 4.382)1(4, 3.521)1 ln C6.105 2 3.521b. M: }} ø 2.8554.905 2 4ln W 2 6.105 5 2.855(ln C 2 4.905)ln W 5 2.855 ln C 2 7.899ln W 5 ln C 2.855 2 7.899W 5 e lnC2.855 2 7.899W 5 e 27.899 ln C2.855eW 5 0.000371C 2.855c. When C 5 68.7: W 5 0.000371C 2.855W 5 0.000371(68.7) 2.855W 5 0.000371(175,594.73)W ø 65.14The weight of a cheetah is about 65.14 kilograms whenthe circumference of its femur is 68.7 millimeters.1(1, 0)4M 5 }4.382 2 0 ø 0.737 2 1ln y 2 4.382 5 0.73(x 2 7)ln y 5 0.73x 2 0.7280.73x 2 0.728y 5 ey 5 e 20.728 (e 0.73 ) xxy 5 0.48(2.08) xb. By graphing the number of non-business users overtime, you can see the graph of the original functionclosely resembles a line more than the graphs of both(x, ln y) and (ln x, ln y).195 2 97M 5 } ø 32.75 2 2y 2 195 5 32.7(x 2 5)y 5 32.7x 1 31.5Copyright © by McDougal Littell, a division of Houghton Mifflin Company.456Algebra 2Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.Chapter 7, continued34. a. lnP 5.004 5.455 5.872 6.265 6.633 6.981lnT 4.094 4.248 4.382 4.500 4.605 4.7001ln P(5.455, 4.248)1(6.633, 4.605)ln T4.605 2 4.248b. M 5 }}6.633 2 5.455 ø 0.30ln T 2 4.605 5 0.30(ln P 2 6.633)c. When P 5 620:ln T 5 0.3ln P 1 2.615ln T 5 ln P 0.3 1 2.615T 5 e ln P0.3 1 2.615T 5 e 2.615 ln P0.3p eT 5 13.6672P 0.3T 5 13.6672(620) 0.3 ø 13.6672(6.8820) ø 94The water boils at a temperature of about 948Cwhen the atmospheric pressure is 620 millimeters ofmercury.35. a. x 10 20 30 40 50 60ln y 2.303 2.485 2.708 3.219 3.689 4.6051ln y10 xb. ln x 2.303 2.996 3.401 3.689 3.912 4.094ln y 2.303 2.485 2.708 3.219 3.689 4.6051ln y1 ln xc. Because the scatter plot from part (a) appears to bemore linear, an exponential model is better fit for theoriginal data.36.3.6 2 2.485d. M 5 } ø 0.04545 2 20ln y 2 2.485 5 0.045(x 2 20)ln y 5 0.045x 1 1.5850.045x 1 1.585y 5 ey 5 e 1.585 (e 0.045 ) xy 5 4.88(1.05) xWhen x 5 80: y 5 4.88(1.05) 80y 5 4.88(49.56)y 5 241.85The linear point for an 80-year-old person is about242 cm.x 0 2 4 6 8ln y 5.147 4.883 4.700 4.522 4.431x 10 12 14ln y 4.357 4.317 4.2771ln y(2, 4.883)2(12, 4.317)4.317 2 4.883M 5 }} ø 20.0612 2 2ln y 2 4.883 5 20.06(x 2 2)ln y 5 20.06x 1 5.003Mixed Review20.06x 1 5.003y 5 ey 5 e 5.003 (e 20.06 ) xy 5 148.86(0.94) x37. y 5 kx when x 5 6 and y 5 4848 5 k(6)8 5 ky 5 8x38. y 5 kx When x 5 27 and y 5 2828 5 k(27)24 5 ky 5 24x39. y 5 kx When x 5 10 and y 5 66 5 k(10)0.6 5 ky 5 0.6xxAlgebra 2Worked-Out Solution Key457
Chapter 7, continued40. y 5 kx when x 5 35 and y 5 1515 5 k(35)3}7 5 ky 5 }3 7 x41. y 5 kx when x 5 0.3 and y 5 1.21.2 5 k(0.3)4 5 ky 5 4x42. y 5 kx when x 5 12 and y 5 1515 5 k(12)5}4 5 ky 5 }5 4 x43.y44.52. 2 ln 6 2 3 ln x 5 ln 6 2 2 ln x 3 5 ln 36 2 ln x 3 5 ln 1}36 x 2 3 3 1.9 1 5 0 1353. log 57 1 6 log 5x 2 log 53 5 log 57 1 log 5x 6 2 log 535 log 57 p x 6 2 log 535 log 5 1 }7x63 254. log 8 2 2 log 2 1 4 log x 5 log 8 2 log 2 2 1 log x 45 log 8 2 log 4 1 log x 45 log }8 1 log x445 log 2 1 log x 4 5 log 2x 455.Cost per Number ofPizza cost1 topping p toppings ≤(dollars)(dollar/topping) (toppings)11 1 2 p x ≤ 1811 1 2x ≤ 18y2x ≤ 7x ≤ 3.5You can order at most 3 toppings if you have $18.11Quiz 7.6–7.7 (p. 536)21x21x1. 2 x 1 1 5 16 x 1 2Domain: all real numbers Domain: all real numbers2 x 1 1 5 (2 4 ) x 1 2Range: y > 0 Range: y > 02 x 1 1 5 2 4x 1 845.f(x)46.yy 5 e 2xx 1 1 5 4x 1 827 5 3x5f(x) 5 2e 2x 1 1(0, 1)1 (1, 0.37)(0, 3)2 }72 x3 5 x(21, 1.27) (0, 2)Check: 2 x 1 1 5 16 x 1 2(21, 0.27) 1 x(0, 23)(1, 23.63)2 (27/3) 1 1 0 16 (27/3) 1 2f(x) 5 2e 2xy 5 e 2x 2 42 24/3 0 16 21/3Domain: all real numbers Domain: all real numbers2 24/3 0 (2 4 ) 21/3Range: y > 1 Range: y > 242 24/3 5 2 24/3 ✓47.y48.g(x)2. e 2x 5 4y 5 2e xg(x) 5 0.5e x 1 1 1 32x 5 ln 44(0, 4.36)x 5 2ln 43x ø 21.39(0, 2) (2, 2)(21, 3.5)(21, 0.73)(1, 1.36) Check: e 2x 5 4(1, 0.73)g(x) 5 0.5e xy 5 2e x 2 2 1xe ln 4 0 4(0, 0.5) 1 x4 5 4 ✓Domain: all real numbers Domain: all real numbers 3. 3 2x 1 5 5 13Range: y > 0 Range: y > 33 2x 5 849. 3 log 74 2 log 78 5 log 74 3 2 log 78 5 log 764 2 log 82x 5 log 38645 log 7}8 5 log 7 8x 5 }log 82 log 350. 2 log 5 1 log 4 5 log 5 2 1 log 4 5 log 25 1 log 4x ø 0.955 log 25 p 4 5 log 100 5 2Check: 3 2x 1 5 5 1351. ln x 1 9 ln y 5 ln x 1 ln y 9 5 ln xy 93 2(0.95) 1 5 0 138.06 1 5 0 1313 ø 13 ✓Amountyou have(dollars)Copyright © by McDougal Littell, a division of Houghton Mifflin Company.458Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.4. 3 x 1 1 2 5 5 103 x 1 1 5 15x 1 1 5 log 315log 15x 5 }log 3 2 1x ø 1.46Check: 3 x 1 1 2 5 5 103 1.46 1 1 2 5 0 103 2.46 2 5 0 1014.92 2 5 0 1010 0 10 ✓5. log 4(4x 1 7) 5 log 411x4x 1 7 5 11x7 5 7x1 5 xCheck: log 4(4x 1 7) 5 log 411xlog 4(4(1) 1 7) 0 log 411(1)6. ln (3x 2 2) 5 ln 6x3x 2 2 5 6x22 5 3x2 2 }3 5 xlog 411 5 log 411 ✓Check: ln (3x 2 2) 5 ln 6xln 1312 }2 32 2 22 0 ln 612 }2 32ln (24) 5 ln (24) ✓Because ln 24 is not defined, there is not solution.7. log 3x 5 21 8. 6 ln x 5 30x 5 3 21 ln x 5 5x 5 1 }3x 5 e 5Check: log 3x 5 21 x ø 148.41log 3 1 }1 032 21 Check: 6 ln x 5 309. log 2(x 1 4) 5 5log }1 3}log 3 0 21 6 ln e 5 0 3021 5 21 ✓ 6(5) 0 30x 1 4 5 2 5x 1 4 5 32x 5 28Check: log 2(x 1 4) 5 5log 2(28 1 4) 0 5log 232 0 5log 32}log 2 0 55 5 5 ✓30 5 30 ✓10. (1, 5): 5 5 ab 1 → a 5 5 }b(2, 30): 30 5 ab 230 5 1 }5 b2 b 230 5 5b6 5 ba 5 }5 b 5 }5 6So, an equation is y 5 5 }6 p 6x11. (1, 4): 4 5 ab 1 → a 5 4 }b(2, 32): 32 5 ab 232 5 1 }4 b2 b 232 5 4b8 5 ba 5 }4 b 5 }4 8 5 }1 2So, an equation is y 5 1 }2 p 8x12. (2, 15): 15 5 ab 2 → a 5 15 }b 2(3, 45): 45 5 ab 345 5 1 15 }b 2 2 b345 5 15b3 5 ba 5 }15 b 5 15 2 }3 2 5 }5 3So, an equation is y 5 5 }3 p 3x13. (4, 8): 8 5 a(4) b → a 5 8 }4 b(9, 23): 23 5 a(9) b23 5 1 8 }4 b 2 9b23 5 81 }9 42 b23}8 5 1 }9 42 blog 9/423}8 5 blog }238}log }9 5 b41.30 ø ba 5 }84 ø 8 b } ø 1.321.34So, an equation is y 5 1.32x 1.3Algebra 2Worked-Out Solution Key459
Chapter 7, continued14. (3, 12): 12 5 a(3) b → a 5 }123 b(10, 36): 36 5 a(10) b36 5 1 12 }3 b 2 10b36 5 121 }10 3 2 b3 5 1 }10 3 2 blog 10/33 5 blog 3}log }10 5 b30.91 ø ba 5 }123 ø 12 b }3 5 4.420.91So, an equation is y 5 4.42x 0.9115. (5, 4): 4 5 a(5) b → a 5 }45 b(11, 51): 51 5 a(11) bMixed Review of Problem Solving (p. 537)1. a.x 1 2 3 4 5ln y 5.986 6.031 6.071 6.109 6.138ln y11xb. ln x 0 0.693 1.099 1.386 1.609ln y 5.986 6.031 6.071 6.109 6.138ln y51 5 1 4 }5 b 2 11b51 5 41 }11 5 2 b51}4 5 1 }11 5 2 blog 11/551}4 5 blog }514}log }11 5 b53.23 ø ba 5 }45 ø 4b } ø 0.023.235So, an equation is y 5 0.02x 3.2316. When y 5 15: y 5 0.51(1.46) x15 5 0.51(1.46) x15}0.51 5 1.46xlog 15 1.4}6 5 x 0.51log }150.51}log 1.46 5 x8.94 ø xA cod is about 9 years old when it weighs 15 kilograms.10.2ln xc. The graph from part (a) is more linear, so anexponential model will be a better fit.6.138 2 6.031M 5 }} ø 0.0365 2 2ln y 2 6.138 5 0.036(x 2 5)ln y 5 0.036x 1 5.9580.036x 1 5.958y 5 ey 5 e 5.958 p (e 0.036 ) xy 5 386.84(1.04) xd. When x 5 8: y 5 386.84(1.04) 8 ø 386.84(1.369)ø 529.58The total expenditures in 2005 will be about 530billion dollars.}E42. a. 1200 log 2}C4 2 1200 log G4}2 E4 5 1200 log 21 E4C4}G4} 2E4b. C4 5 264; E4 5 330; G4 5 3965 1200 log 2 1 (E4)2 }C4 p G421200 log 2 1 }(E4)2330C4 p G42 5 1200 log 2 1 }2264 p 39625 1200 log 2 1 }25 1200 log }25 24242 5 }log 2Copyright © by McDougal Littell, a division of Houghton Mifflin Company.ø 1200(0.0589) 5 70.68The difference in the number of cents from C4 to E4and the number of cents from E4 to G4 is about 70.68cents.460Algebra 2Worked-Out Solution Key
Chapter 7, continued3. Sample answer:Known point: (2, 7) Another point: (4, 112)(2, 7): 7 5 ab 2 → a 5 }7 b 2(4, 112): 112 5 ab 44. A logarithm with base e is called a natural logarithm.5. y 5 (1.4) x is an exponential function because it has theform y 5 ab x .6.y7.yCopyright © by McDougal Littell, a division of Houghton Mifflin Company.112 5 1 7 }b 2 2 b4112 5 7b 216 5 b 24 5 ba 5 }7 b 5 7 2 }4 5 7 2 }16An equation is y 5 }7 16 p 4x .4. When you look at the graphs of (x, ln y) and(ln x, ln y), graph of (ln x, ln y) appears to be more linearthan the graph of (x, ln y). So, a power model wouldbetter fit the data.5. a. When N 5 0.05: N 5 ln (E 1 1)0.05 5 ln (E 1 1)e 0.05 5 E 1 1e 0.05 2 1 5 Eb. When N 5 0.1: N 5 ln (E 1 1)0.1 5 ln (E 1 1)e 0.1 5 E 1 1e 0.1 2 1 5 EE from part (b)c. }}E from part (a) 5 e0.1 2 1}e 0.05 2 1e 0.1 2 1d. } 0 e 0.05 1 1e 0.05 2 1(e 0.05 2 1) 1 e0.01 2 1}e 0.05 2 12 0 (e 0.05 2 1)(e 0.05 1 1)e 0.1 2 1 0 e 2(0.05) 1 e 0.05 2 e 0.05 2 1e 0.1 2 1 5 e 0.1 2 1 ✓6. When P 5 4000, r 5 0.02, andA 5 4000 1 1000 5 5000:A 5 Pe rt5000 5 4000e (0.02)t1.25 5 e 0.02tln 1.25 5 0.02t0.22 ø 0.02t11 ø tIt will take about 11 years to earn $1000 interest.Chapter 7 Review (pp. 539–542)1. The asymptote of the function y 5 221 }1 42 x 1 1 1 5is y 5 5.2. The decay factor in the model y 5 7.2(0.89) x is 0.89.3. Sample answer: log by represents the logarithm of y withbase b, and it is equal to x if and only if b x 5 y.212xDomain: all real numbers Domain: all real numbersRange: y > 0 Range: y > 08.3 f(x)f(x) 5 23 ? 4 x( 21, 24 )( 22, 2114 )(21, 25)f(x) 5 23 ? 4 x 1 1 2 211(0, 23)Domain: all real numbersRange: y < 229. When P 5 1500, r 5 0.07n 5 365, and t 5 2:A 5 P 11 1 } n r 2 ntA 5 150011 1 }0.07365 2 365 p 25 15001 }365.07365 2 730ø 1725.39The balance after 2 years is $1725.39.y10.11.211xx(0, 23)21yy 5(0, 1)1y 5211( 3)x1( 1,3 )( 1, 2113 )1( ) x 3 2 4Domain: all real numbers Domain: all real numbersRange: y > 0 Range: y > 2412.f(x)13.yf(x) 5 2(0.8) x 2 1 1 3(1, 5)(2, 4.6)(0, 2) f(x) 5 2(0.8) x(1, 1.6)211xDomain: all real numbers Domain: all real numbersRange: y > 3 Range: y > 0211xxxAlgebra 2Worked-Out Solution Key461
Chapter 7, continued14.y 5 e x(1, 2.72)(0, 1)2121y(3, 2.72)(2, 1)y 5 e x 2 2Domain: all real numbersRange: y > 015.f(x)(22, 7)21f(x) 5 e 20.4(x 1 2) 1 6(21, 6.67)2(0, 1)(1, 0.67)xf(x) 5 e 20.4xDomain: all real numbersRange: y > 616. When t 5 10: N 5 6.9e 20.0695tN 5 6.9e 20.0695(10)N 5 6.9e 20.695N ø 3.44About 3.44 picograms of nitrogen-13 remain after10 minutes.17. 3 5 5 243, so log 3243 5 5.18. 7 0 5 1, so log 71 5 0.19. 1 }1 62 23 5 216, so log 1/6216 5 23.20. 125 21/3 5 }1 5 , so log 1}125 5 5 2 }1 3 .21. y22.211xxyy 5 log 3 x1(3, 1)21 (1, 0)(3, 23)(1, 24)y 5 log 3 x 2 4Domain: x > 0 Domain: x > 0Range: all real numbers Range: all real numbers23. f(x)f(x) 5 ln(x 2 1) 1 3(5, 4.39)(2, 3)1(4, 1.39)21 (1, 0)f(x) 5ln xxx24. d 5 22.1158 ln a 1 13.669When a 5 25: d 5 22.1158 ln 25 1 13.669d ø 22.1158(3.2189) 1 13.669 ø 6.8585When a 5 50: d 5 22.1158 ln 50 1 13.669d ø 22.1158(3.9120) 1 13.669 ø 5.3920The average diameter of a pupil is about 6.8585millimeters for a 25 year old and and about 5.3920millimeters for a 50 year old.25. log 83xy 5 log 83 1 log 8x 1 log 8y26. ln 10x 3 y 5 ln 10 1 ln x 3 1 ln y 5 ln 10 1 3 ln x 1 ln y27. log 8 }y 4 5 log 8 2 log y4 5 log 8 2 4 log y28. ln 3y }x 5 5 ln 3y 2 ln x5 5 ln 3 1 ln y 2 5 ln x29. 3 log 74 1 log 76 5 log 74 3 1 log 76 5 log 764 1 log 765 log 764 p 6 5 log 738430. ln 12 2 2 ln x 5 ln 12 2 ln x 2 5 ln 12 }x 231. 2 ln 3 1 5 ln 2 2 ln 8 5 ln 3 2 1 ln 2 5 2 ln 85 ln 9 1 ln 32 2 ln 85 ln 9 p 32 2 ln 85 ln 288 2 ln 8 5 ln }288 5 ln 36832. 5 x 5 32log 55 x 5 log 532x 5 log 532log 32x 5 }log 5 5 2.153Check: 5 x 5 325 2.153 0 325 2.153 ø 32, 32 5 32 ✓33. log 3(2x 2 5) 5 23 log 3 (2x 2 5) 5 3 22x 2 5 5 92x 5 14x 5 7Check: log 3(2x 2 5) 5 2log 3(2 p 7 2 5) 0 2log 39 0 23 2 5 9, log 39 5 2 ✓Copyright © by McDougal Littell, a division of Houghton Mifflin Company.Domain: x > 1Range: all real numbers462Algebra 2Worked-Out Solution Key
Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.34. ln x 1 ln (x 1 2) 5 3ln [x(x 1 2)] 5 3e ln [x(x 1 2)] 5 e 3x(x 1 2) 5 e 3x 2 1 2x 2 e 3 5 0x 5 226 Ï } 4 2 4(2e 3 )}}2ø 3.59, 25.59Check: ln x 1 ln (x 1 2) 5 3ln (25.59) 1 ln (25.59 1 2) 0 3ln (25.59) 1 ln (23.59) 0 3ø 226 Ï} 84.34}2Because ln (25.59) and ln (23.59) are not defined, 25.59is not a solution.ln x 1 ln (x 1 2) 5 3ln 3.59 1 ln (3.59 1 2) 0 3ln 3.59 1 ln 5.59 0 3ln 3.59 p 5.59 0 3ln 20.07 0 33 ø 3 ✓The solution is about 3.59.35. (3, 8): 8 5 ab 3 → a 5 8 }b 3(5, 2): 2 5 ab 52 5 1 8 }b 3 2 b52 5 8b 21}4 5 b21}2 5 ba 5 }8 b 5 83 }1 }22 1 5 64 3So, an equation is y 5 64 p 1 }1 22 x .36. (22, 2): 2 5 ab 22 → a 5 2 }b 22(1, 0.25): 0.25 5 ab 10.25 5 1 2 }b 22 2 b10.25 5 2b 30.125 5 b 30.5 5 ba 5 }2 b 5 222 }225 0.50.5So, an equation is y 5 0.5 p 0.5 x .37. (2, 9): 9 5 ab 2 → a 5 9 }b 238.(4, 324): 324 5 ab 4324 5 1 9 }b 2 2 b4324 5 9b 236 5 b 26 5 ba 5 }9 b 5 9 2 }36 5 }1 4So, an equation is y 5 1 }4 p 6x .ln x 0 0.693 1.099 1.386 1.609 1.792ln y 3.332 3.850 4.159 4.369 4.543 4.673M:0.5ln y(0, 3.332)0.2(1.792, 4.673)4.673 2 3.332}} ø 0.751.792 2 0ln xln y 2 4.673 5 0.75(ln x 2 1.792)ln y 5 0.75 ln x 1 3.329ln y 5 ln x 0.75 1 3.329y 5 e ln x0.75 1 3.329y 5 e 3.329 ln x0.75p ey 5 27.91x 0.75Chapter 7 Test (p. 543)1.212yx2.y 5 2 ? 4 x 2 22(0, 2)y 5 2 ? 4 x21y(1, 8) (3, 8)(2, 2)Domain: all real numbers Domain: all real numbersRange: y > 0 Range: y > 0xAlgebra 2Worked-Out Solution Key463
Chapter 7, continued3.f(x) 5 25 ? 2 x 1 3 1 31( 24,2 ) 1(23, 22)f(x) 5 25 ? 2 x f(x)1 x5( 21, 22(0, 25)4.)Domain: all real numbers Domain: all real numbersRange: y < 3 Range: y > 05.y6.g(x)(22, 2) (0, 2)2( 21,3)( )11 x 1 2y 5 2 31 xy 5 2 31( )2( 1,3)x21(0, 3)(0, 1)211yg(x) 581,3( )2( 1,3)x2( ) x 3 1 2( )12 x xg(x) 5 3Domain: all real numbers Domain: all real numbersRange: y > 0 Range: y > 27.y8.y212x(0, 2.5)(1, 1.52)2121y 5 2.5e 20.5x 1 1(0, 3.5)(1, 2.52)y 5 2.5e 20.5x xDomain: all real numbers Domain: all real numbersRange: y > 0 Range: y > 19.h(x)( )1h(x) 5 e x34(1, 0.91)1( 0,3)5 21( 1, 23)(2, 21.09) x( )1h(x) 5 e x 2 1 2 23Domain: all real numbersRange: y > 2210. 5 2 5 25, so log 525 5 2. 11. 2 25 5 }1 32 , so log 1}2 32 5 25.12. 6 0 5 1, so log 61 5 0.13.211yx14.yy 5 ln x(1, 0)1(2, 0.69)21x(2, 22.31)(1, 23)y 5 ln x 2 3Domain: x > 0 Domain: x > 0Range: all real numbers Range: all real numbers15.f(x)f(x) 5 log(x 1 3) 1 2(21, 2.30)(22, 2)1(2, 0.30)21 (1, 0)f(x) 5log xxDomain: x > 23Range: all real numbers16. 2 ln 7 2 3 ln 4 5 ln 7 2 2 ln 4 3 5 ln 49 2 ln 64 5 ln }496417. log 43 1 5 log 42 5 log 43 1 log 42 55 log 43 1 log 4325 log 43 p 32 5 log 49618. log 5 1 log x 2 2 log 3 5 log 5 1 log x 2 log 3 25 log 5 1 log x 2 log 95 log 5 p x 2 log 9 5 log }5x9log 5019. log 550 5 }log 5 ø 2.431 20. log log 2323 5 }6 log 6 ø 1.750log 4521. log 945 5 }log 9 ø 1.73222. 7 2x 5 30 23. 3 log (x 2 4) 5 6log 77 2x 5 log 730 log (x 2 4) 5 22x 5 log 730 10 log (x 2 4) 5 10 2log 30x 5 }2 log 7x 2 4 5 100x ø 0.8740 x 5 104Check: 7 2x 5 30 Check: 3 log (x 2 4) 5 67 2(0.8740) 0 30 3 log (104 2 4) 0 67 1.748 0 30 3 log 100 0 630 ø 30 ✓ 3 p 2 0 66 5 6 ✓24. log 4x 1 log 4(x 1 6) 5 2log 4[x(x 1 6)] 5 24 log 4[x(x 1 6)] 5 4 2x(x 1 6) 5 16x 2 1 6x 5 16x 2 1 6x 2 16 5 0(x 1 8)(x 2 2) 5 0x 5 28 or x 5 2Check: log 4x 1 log 4(x 1 6) 5 2log 4(28) 1 log 4(28 1 6) 0 2log 4(28) 1 log 4(22) 0 2Because log 4(28) and log 4(22) are not defined, 28 is nota solution.log 4x 1 log 4(x 1 6) 5 2log 42 1 log 4(2 1 6) 0 2log 42 1 log 48 0 2log 416 0 2Because 4 2 5 16, log 416 5 2. ✓The solution is 2.Copyright © by McDougal Littell, a division of Houghton Mifflin Company.464Algebra 2Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.Chapter 7, continued25. (21, 48): 48 5 ab 21 → a 5 48 }b 21(2, 6): 6 5 ab 26 5 1 48 }b 21 2 b26 5 48 p b 31}8 5 b31}2 5 ba 5 }48 b 5 4821 }1 }1 215 2422So, an equation is y 5 241 }1 22 x .26. (3, 8): 8 5 a(3) b → a 5 8 }3 b(6, 15): 15 5 a(6) b15 5 1 8 }3 b 2 6b15 5 8 p 2 b15}8 5 2b15log }8}log 2 5 b0.907 ø ba 5 }83 ø 8b } ø 2.9540.9073So, an equation is y 5 (2.954)x 0.907 .27. H 5 a(1 1 r) tThe initial amount is a 5 62 and the percent increaseis r 5 0.0485. So, the exponential growth model isH 5 62(1.0485) t .28. When P 5 2500, r 5 0.035, and t 5 8: A 5 Pe rtA 5 2500e 0.035(B) 5 2500e 0.28 ø 3307.82The balance after 8 years is $3307.82.29. a. ln x 21.609 1.609 2.398 2.996 3.807ln y 22.303 20.643 20.288 0 0.405211ln yln x0.405 2 (22 p 203)b. M 5 }}3.807 2 (21 p 609) 5 0.5ln y 2 0.405 5 0.5(ln x 2 3.807)When x 5 120:ln y 5 0.5ln x 2 1.4985ln y 5 ln x 0.5 2 1.4985y 5 e ln x0.5 2 1.4985y 5 e 21.4985 ln x0.5ey 5 0.2235x 0.5y 5 0.2235(120) 0.5 ø 0.2235(10.9545) → y ø 2.4483The speed of the current needed to carry a particle witha diameter of 120 millimeters downstream is about 2.45meters per second.Standardized Test Preparation (pp. 544 –545)1. Sample answer: You can eliminate choice C by substituting22 for x. Because taking the square root of a negativenumber yields an imaginary answer, choice C can beeliminated.2. Sample answer: You can eliminate choice A by substituting4 for x and seeing that the answer is greater than 17.e 8 > 2 8 > 2 5 5 32 > 173. Sample answer: you can eliminate choice B by substituting22 for x in the function. Because this does not make thefunction undefined, choice B can be eliminated.Standardized Test Practice (pp. 546 –547)1. C; The initial amount was 22,054 and the percentincrease was 0.0425. So, the exponential growth modelis y 5 22,054(1.0425) t .2. A; When P 5 1000, r 5 0.03, and t 5 2: A 5 P 11 1 }n2 r ntDaily: n 5 365A 5 1000 1 1 1 0.03 }Monthly: n 5 12365 2 356 p 2 5 10001 }365.03365 2 730 ø 1061.83A 5 1000 1 2 1 0.03 }12 2 12 p 2 5 1000(1.0025) 24 ø 1061.76Daily 2 monthly 5 1061.83 2 1061.76 5 0.07There is $.07 more interest earned if it is compoundeddaily.3. B; y 5 22 p 4 x 1 1when x 5 0 and y 5 21: When x 5 1 and y 5 27:21 0 22 p 4 0 1 1 27 0 22 p 4 1 1 121 0 22 1 1 27 0 28 1 121 5 21 ✓ 27 5 27 ✓4. D; To translate the grpah of y 5 log 3(x 1 2) 2 4 left 1unit, and 1 to x. The new graph is y 5 log 3(x 1 3) 2 4.5. C; 3 log x 1 log 3 5 log x 3 1 log 3 5 log x 3 p 3 5 log 3x 36. B; y 5 290(1.009) t ; t 5 2010 2 2003 5 7y 5 290(1.009) 7 ø 290(1.0647) ø 309The projected population for 2010 is about 309 million.7. A; Ï } 100e 6x 5 Ï } 100 Ï } e 6x 5 10e 3xAlgebra 2Worked-Out Solution Key465
Chapter 7, continued8. D; y 5 2(0.5) x is the only exponential decay functionbecause 0 < 0.5 < 1.9. C; y 5 e 2x 2 3x 5 e 2y 2 3x 1 3 5 e 2yln (x 1 3) 5 ln (e 2y )ln (x 1 3) 5 2yln (x 1 3)} 5 y210. B; log (x 1 3) 1 log x 5 1log [x(x 1 3)] 5 110 log [x(x 1 3)] 5 10 1x(x 1 3) 5 10x 2 1 3x 5 10x 2 1 3x 2 10 5 0(x 1 5)(x 2 2) 5 0x 5 25 or x 5 2Check: log (x 1 3) 1 log x 5 1log (25 1 3) 1 log (25) 0 1log (22) 1 log (25) 0 1Because log 22 and log 25 are not defined, 25 is not asolution.log (x 1 3) 1 log x 5 1log (2 1 3) 1 log 2 0 1log 5 1 log 2 0 1log 10 0 11 5 1 ✓The solution is 2.11. B; y 5 4x 0.5x 5 1 and y 5 4: x 5 4 and y 5 8:4 0 4(1) 0.5 8 0 4(4) 0.54 0 4(1) 8 0 4(2)4 5 4 ✓ 8 5 8 ✓12. When P 5 500, r 5 0.035, and t 5 1: A 5 Pe rtA 5 500e 0.035 p 1 5 500e 0.035 ø 518The amount of interest earned is $518 2 $500 5 $18.13. When y 5 50: y 5 7.7e 0.14x50 5 7.7e 0.14x50}7.7 5 e0.14xln }50 7.7 5 0.14x1.8708 ø 0.14x13.4 5 xThere will be 50,000 bacteria per cubic centimeter afterabout 13.4 hours.14. 9 2x 1 1 5 3 5x 2 1(3 2 ) 2x 1 1 5 3 5x 2 13 4x 1 2 5 3 5x 2 14x 1 2 5 5x 2 13 5 x15. log 272 2 2 log 23 5 log 272 2 log 23 2 5 log 272 2 log 29725 log 2}9 5 log 2 85 log 2 23 5 316. (0, 8): 8 5 ab 0 → a 5 }8 b 5 8 0(2, 2): 2 5 ab 22 5 (8)b 21}4 5 b21}2 5 bAn equation is f(x) 5 8 p 1 }1 22 x .When x 5 5: f (5) 5 8 p 1 }1 22 5 5 }1 417. log 3(5x 1 3) 5 53 log3 (5x 1 3) 5 3 55x 1 3 5 2435x 5 240x 5 4818. When T 05 232, T R5 68, t 5 20, and T 5 110:T 5 (T 02 T R)e 2rt 1 T R110 5 (232 2 68)e 2r p 20 1 6842 5 164e 220r21}82 5 e220rln }21 5 ln e220r82ln }21 82 5 220r21.362 5 220r0.068 5 rThe cooling rate is about 0.068. Sample answer: If youwere in a hurry, you could increase the cooling rate byputting the fudge in the freezer. The value of T Rwoulddecrease and the value of r would increase.19. y 5 37(1 2 0.30) x 5 37(0.70) x ; 0 ≤ x ≤ 11.Sample answer: This domain is how long it takes the movieto gross less than a million dollars a week.20.yThe graph of y 5 2 2x growsmore quickly because the powery 5 2 xis 2 x and 2 x > x.2y 5 2 2xCopyright © by McDougal Littell, a division of Houghton Mifflin Company.21x466Algebra 2Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.Chapter 7, continued21. a.x 0 5 8 11 15ln y 8.700 10.275 11.806 12.525 13.998x 19 23 25 26 30ln y 14.947 15.830 16.067 17.553 18.6443ln y(2, 9.539)5(19, 14.947)14.947 2 9.539b. M 5 }} ø 0.318119 2 2ln y 2 14.947 5 0.3181(x 2 19)ln y 5 0.3181x 1 8.90310.3181x 1 8.9031y 5 exy 5 e 8.9031 p (e 0.3181 ) xy 5 7355(1.37) xc. When x 5 2008 2 1974 5 34: y 5 7355(1.37) xy 5 7355(1.37) 34 ø 327,402,447There will be about 327,400,000 transisters perintegrated circuit in 2008.d. 18 months 5 1.5 yearsWhen x 5 0: y 5 7355(1.37) 0 5 7355When x 5 1.5: y 5 7355(1.37) 1.5 ø 11,794When x 5 3: y 5 7355(1.37) 3 ø 18,912According to the model, Moore’s law does not holdbecause the value of y does not double every 1.5 years.22. a. Sample answer: The data is just short of doublingbecause the equation came from a few specific datapoints, not a generalization.When h 5 10,000: D 5 0.761e 20.0004hD 5 0.0761e 20.0004(10,000) 5 0.0761e 24 ø 0.00139The density of air is about 0.00139 pounds per cubicfoot at 10,000 feet above sea level.b. When P 5 12: P 5 14.7e 20.0004h12 5 14.7e 20.0004h12}14.7 5 e20.0004hln }12 14.7 5 20.0004h20.2029 ø 20.0004h507 ø hThe air pressure is 12 pounds per square inch at about507 feet above sea level.c. }D P 5 }0.0761 ø 0.005 The air density is roughly 0.5%14.7of the air pressure at 608F. Both are the same parentgraph with a different vertical shrink and stretch.d. When h 5 0: P 5 14.7e 20.0004(0) 5 14.7e 0 5 14.7when h 5 80: P 5 14.7e 20.0004(80)5 14.7e 20.032 ø 14.237When h 5 160: P 5 14.7e 20.0004(160)5 14.7e 20.064 ø 13.789P 802 P 0 14.237 2 14.7} 5 }} ø 20.03P 014.7P 1602 P 80 13.789 2 14.237} 5 }} ø 20.03P 8014.237No; the air pressure decreases by about 3% for every 80meter increase in altitude.Algebra 2Worked-Out Solution Key467
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