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
- Page 1 and 2: Chapter 7Copyright © by McDougal L
- Page 4 and 5: Chapter 7, continued37. A 5 2200 1
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- Page 8 and 9: Chapter 7, continuedb. When t 5 50:
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- Page 12 and 13: Chapter 7, continuedd. P(2.75) 5 30
- Page 16 and 17: Chapter 7, continued72. x 1/2 p x 2
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- Page 20 and 21: Chapter 7, continuedlog (100h)b. s(
- Page 22 and 23: Chapter 7, continuedCheck: log 5x 1
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- Page 36 and 37: Chapter 7, continued17. (2, 3): 3 5
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- Page 44 and 45: Chapter 7, continued14.y 5 e x(1, 2
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- Page 48 and 49: Chapter 7, continued8. D; y 5 2(0.5