Math Chapter 2 Logarithmic Functions Answer Key

2.5 Logarithmic Functions 1. 53 = 125 Since ay = x means y = loga x; the equation in logarithmic form is log5 125 = 3: 2. 72 = 49 Since ay = x means y = loga x; the equation in logarithmic form is log7 49 = 2: 3. 34 = 81 The equation in logarithmic form is log3 81 = 4: 4. 27 = 128 Since ay = x means y = loga x; the equation in logarithmic form is log2 128 = 7: 1 9 The equation in logarithmic form is 5. 3¡2 = log3 6. 1 = ¡2: 9 μ ¶¡2 4 25 = 5 16 Since ay = x means y = loga x; the equation in logarithmic form is log4=5 25 = ¡2: 16 7. log2 32 = 5 Since y = loga x means ay = x; the equation in exponential form is 25 = 32: 8. log3 81 = 4 Since y = loga x means ay = x; the equation in exponential form is 34 = 81: Chapter 2 1 = ¡1 e The equation in exponential form is 9. ln 1 e¡1 = : e 1 = ¡3 8 The equation in exponential form is 10. log2 1 =x 81 1 3x = 81 18. log3 3x = 3¡4 x = ¡4 q 19. log2 3 14 = x μ ¶1=3 1 2 = 4 μ ¶1=3 1 2x = 22 x 1 2¡3 = : 8 11. log 100,000 = 5 log10 100,000 = 5 105 = 100,000 2x = 2¡2=3 x=¡ When no base is written, log10 is understood. 12. log 0:001 = ¡3 log10 0:001 = ¡3 10¡3 = 0:001 When no base is written, log10 is understood. 13. Let log8 64 = x: Then, 8x = 64 8 x = 82 x = 2: Thus, log8 64 = 2: 14. Let log9 81 = x: r 20. log8 4 15. log4 64 = x 4x = 64 4x = 43 x=3 16. log3 27 = x 3x = 27 3x = 33 x=3 17. log2 1 =x 16 1 2x = 16 2x = 2¡4 x = ¡4 2 3 1 =x 2 r 8x = 4 1 = 2 μ ¶1=4 1 2 (23 )x = 2¡1=4 1 4 1 x=¡ 12 3x = ¡ 21. ln e = x Recall that ln means loge : Then, 9x = 81 9 x = 92 x = 2: Thus, log9 81 = 2: NONLINEAR FUNCTIONS ex = e x=1 22. ln e3 = x Recall that ln means loge : ex = e3 x=3 23. ln e5=3 = x ex = e5=3 x= 5 3 24. ln 1 = x ex = 1 ex = e0 x=0 25. The logarithm to the base 3 of 4 is written log3 4: The subscript denotes the base. Section 2.5 Logarithmic Functions 27. log5 (3k) = log5 3 + log5 k 28. log9 (4m) = log9 4 + log9 m 29. log3 3p 5k = log3 3p ¡ log3 5k = (log3 3 + log3 p) ¡ (log3 5 + log3 k) = 1 + log3 p ¡ log3 5 ¡ log3 k 30. log7 36. logb (9b2 ) = logb 9 + logb b2 = logb 32 + logb b2 = 2 logb 3 + 2 logb b = 2c + 2(1) = 2c + 2 ln 30 ln 5 3:4012 ¼ 1:6094 37. log5 30 = 15p 7y = log7 15p ¡ log7 7y = (log7 15 + log7 p) ¡ (log7 7 + log7 y) = log7 15 + log7 p ¡ log7 7 ¡ log7 y = log7 15 + log7 p ¡ 1 ¡ log7 y ¼ 2:113 ln 210 ln 12 ¼ 2:152 38. log12 210 = p 3 5 31. ln p 3 6 p p = ln 3 5 ¡ ln 3 6 = ln 3 ¢ 51=2 ¡ ln 61=3 = ln 3 + ln 51=2 ¡ ln 61=3 = ln 3 + 1 1 ln 5 ¡ ln 6 2 3 p 935 32. ln p 4 3 p p = ln 9 3 5 ¡ ln 4 3 = ln 9 ¢ 51=3 ¡ ln 31=4 = ln 9 + ln 51=3 ¡ ln 31=4 = ln 9 + 1 1 ln 5 ¡ ln 3 3 4 33. logb 32 = logb 25 = 5 logb 2 = 5a 34. logb 18 = logb (2 ¢ 9) = logb (2 ¢ 32 ) = logb 2 + logb 32 = logb 2 + 2 logb 3 = a + 2c 35. logb 72b = logb 72 + logb b = logb 72 + 1 = logb 23 ¢ 33 + 1 = logb 23 + logb 32 + 1 = 3 logb 2 + 2 logb 3 + 1 = 3a + 2c + 1 39. log1:2 0:95 = ln 0:95 ln 1:2 ¼ ¡0:281 ln 0:12 ln 2:8 ¼ ¡2:059 40. log2:8 0:12 = 41. logx 36 = ¡2 x¡2 = 36 ¡2 ¡1=2 = 36¡1=2 (x ) 1 x= 6 42. log9 27 = m 9m = 27 2 m (3 ) = 33 32m = 33 2m = 3 m= 3 2 43. log8 16 = z 8z = 16 3 z (2 ) = 24 23z = 24 3z = 4 z= 4 3 Chapter 2 logy 8 = 44. (y 3 4 y3=4 = 8 ) = 84=3 y = (81=3 )4 = 24 = 16 3=4 4=3 45. logr 5 = 1 2 NONLINEAR FUNCTIONS 50. log (x + 5) + log (x + 2) = 1 log [(x + 5)(x + 2)] = 1 (x + 5)(x + 2) = 101 x2 + 7x + 10 = 10 x2 + 7x = 0 x(x + 7) = 0 x = 0 or x = ¡7 x = ¡7 is not a solution of the original equation because if x = ¡7; x + 5 and x + 2 would be negative, and the domain of y = log x is (0; 1): Therefore, x = 0: 1=2 =5 r (r1=2 )2 = 52 r = 25 46. log4 (5x + 1) = 2 42 = 5x + 1 16 = 5x + 1 5x = 15 x=3 47. log5 (9x ¡ 4) = 1 51 = 9x ¡ 4 9 = 9x 1=x 51. log3 (x ¡ 2) + log3 (x + 6) = 2 log3 [(x ¡ 2)(x + 6)] = 2 (x ¡ 2)(x + 6) = 32 x2 + 4x ¡ 12 = 9 x2 + 4x ¡ 21 = 0 (x + 7)(x ¡ 3) = 0 x = ¡7 or x = 3 x = ¡7 does not check in the original equation. The only solution is 3: 52. log3 (x2 + 17) ¡ log3 (x + 5) = 1 48. log4 x ¡ log4 (x + 3) = ¡1 x = ¡1 log4 x+3 x 4¡1 = x+3 x 1 = 4 x+3 4x = x + 3 3x = 3 x=1 log3 x2 + 17 =1 x+5 x2 + 17 31 = x+5 3x + 15 = x2 + 17 0 = x2 ¡ 3x + 2 0 = (x ¡ 1)(x ¡ 2) x = 1 or x = 2 53. log2 (x2 ¡ 1) ¡ log2 (x + 1) = 2 log2 49. log9 m ¡ log9 (m ¡ 4) = ¡2 m = ¡2 log9 m¡4 m 9¡2 = m¡4 m 1 = 81 m¡4 m ¡ 4 = 81m ¡4 = 80m ¡0:05 = m This value is not possible since log9 (¡0:05) does not exist. Thus, there is no solution to the original equation. x2 ¡ 1 =2 x+1 x2 ¡ 1 22 = x+1 (x ¡ 1)(x + 1) 4= x+1 4 =x¡1 x=5 54. 2x = 6 ln 2x = ln 6 x ln 2 = ln 6 x= ln 6 ¼ 2:585 ln 2 Section 2.5 55. Logarithmic Functions 5x = 12 x log 5 = log 12 x= 60. log 12 log 5 5(0:10)x = 4(0:12)x ln[5(0:10)x ] = ln[4(0:12)x ] ln 5 + x ln 0:10 = ln 4 + x ln 0:12 x(ln 0:12 ¡ ln 0:10) = ln 5 ¡ ln 4 ¼ 1:544 x= ln 5 ¡ ln 4 ln 0:12 ¡ ln 0:10 ¼ 1:224 56. ek¡1 = 6 ln ek¡1 = ln 6 (k ¡ 1) ln e = ln 6 61. ln 6 ln e ln 6 k¡1 = 1 k¡1 = x= 62. f(x) = log (5 ¡ x) 5¡x >0 ¡x > ¡5 x<5 e2y = 15 ln e2y = ln 15 2y ln e = ln 15 2y(1) = ln 15 The domain of f is x < 5: 63. f(x) = ln (x2 ¡ 9) Since the domain of f(x) = ln x is (0; 1); the domain of f (x) = ln (x2 ¡ 9) is the set of all real numbers x for which ln 15 y= 2 ¼ 1:354 58. x2 ¡ 9 > 0: To solve this quadratic inequality, …rst solve the corresponding quadratic equation. 2e5a+12 = 10 e5a+12 = 5 ln e5a+12 = ln 5 (5a + 12) ln e = ln 5 5a + 12 = ln 5 a= x2 ¡ 9 = 0 (x + 3)(x ¡ 3) = 0 x+3 =0 or x ¡ 3 = 0 x = ¡3 or x =3 These two solutions determine three intervals on the number line: (¡1; ¡3); (¡3; 3); and (3; 1): ln 5 ¡ 12 5 If x = ¡4; (¡4 + 2)(¡4 ¡ 2) > 0: If x = 0; (0 + 2)(0 ¡ 2) ¦ 0: If x = 4; (4 + 2)(4 ¡ 2) > 0: a ¼ ¡2:078 59. ln 2 ¡ ln 1:5 ln 1:05 ¡ ln 1:01 ¼ 7:407 k = 1 + ln 6 ¼ 2:792 57. 1:5(1:05)x = 2(1:01)x ln[1:5(1:05)x ] = ln[2(1:01)x ] ln 1:5 + x ln 1:05 = ln 2 + x ln 1:01 x(ln 1:05 ¡ ln 1:01) = ln 2 ¡ ln 1:5 10e3z¡7 = 100 ln 10e3z¡7 = ln 100 ln 10 + ln e3z¡7 = ln 100 ln 10 + (3z ¡ 7) ln e = ln 100 3z ¡ 7 = ln 100 ¡ ln 10 3z = ln 100 ¡ ln 10 + 7 z= ln 100 ¡ ln 10 + 7 3 ¼ 3:101 The domain is x < ¡3 or x > 3; which is written in interval notation as (¡1; ¡3) [ (3; 1): 64. log A ¡ log B = 0 log A =0 B A = 100 = 1 B A=B A¡B =0 Chapter 2 Thus, solving log A ¡ log B = 0 is equivalent to solving A ¡ B = 0: x 65. Let m = loga ; n = loga x; and p = loga y: y x Then am = ; an = x; and ap = y: y Substituting gives For r = 0:08; we use the rule of 72. 72 = 9 years 100(0:08) 68. (a) t = an x = p = an¡p : y a am = NONLINEAR FUNCTIONS t= So m = n ¡ p: Therefore, ln 2 ln (1 + r) ln 2 ln (1 + 0:07) t ¼ 10:24 It will take 11 years for the compound amount to be at least double. ln 3 (b) t = ln (1 + 0:07) x = loga x ¡ loga y: y loga 66. Let m = loga xr and n = loga x: Then, am = xr and an = x: Substituting gives t ¼ 16:24 It will take 17 years for the compound amount to at least triple. am = xr = (an )r = anr : Therefore, m = nr; or (c) The rule of 72 gives loga x = r loga x: r 72 = 10:29 100(0:07) 67. From Example 7, the doubling time t in years when m = 1 is given by t= ln 2 : ln (1 + r) years as the doubling time. 69. A = P ert 1200 = 500er¢14 2:4 = e14r ln(2:4) = ln e14r ln(2:4) = 14r (a) Let r = 0:03: ln 2 ln 1:03 = 23:4 years t= ln(2:4) =r 14 (b) Let r = 0:06: t= ln 2 ln 1:06 0:0625 ¼ r The interest rate should be 6.25%. = 11:9 years (c) Let r = 0:08: 70. r (sec) 0:001 0:02 0:05 0:08 0:12 ln 2 ln(1 + r) 693:5 35 14:2 9:01 6:12 (d) Since 0:001 · 0:03 · 0:05; for r = 0:03; we use the rule of 70. 70 100r 700 35 14 8:75 5:83 70 70 = = 23:3 years 100r 100(0:03) 72 100r 720 36 14:4 9 6 t= ln 2 ln 1:08 = 9:0 years Since 0:05 · 0:06 · 0:12; for r = 0:06; we use the rule of 72. 72 72 = = 12 years 100r 100(0:06) For 0:001 · r < 0:05; the Rule of 70 is more accurate. For 0:05 < r · 0:12; the Rule of 72 is more accurate. At r = 0:05; the two are equally accurate. Section 2.5 Logarithmic Functions 71. After x years at Humongous Enterprises, your salary would be 45,000 (1 +0.04)x or 45,000 (1.04)x : After x years at Crabapple Inc., your salary would be 30,000 (1 + 0:06)x or 30,000 (1.06)x : First we …nd when the salaries would be equal. 45,000(1:04)x = 30,000(1.06)x 30,000 (1:04)x = (1:06)x 45,000 ¶x μ 2 1:04 = 1:06 3 μ μ ¶ ¶x 1:04 2 log = log 1:06 3 μ ¶ μ ¶ 1:04 2 x log = log 1:06 3 μ ¶ 2 log 3 μ ¶ x= 1:04 log 1:06 74. H = ¡[P1 ln P1 + P2 ln P2 + P3 ln P3 + P4 ln P4 ] H = ¡[0:521 ln 0:521 + 0:324 ln 0:324 + 0:081 ln 0:081 + 0:074 ln 0:074] H = 1:101 75. (a) 3 species, 72. If the number N is proportional to m¡0:6 ; where m is the mass, then N = km¡0:6 ; for some constant of proportionality k: Taking the common log of both sides, we have log N = log(km¡0:6 ) = log k + log m¡0:6 = log k ¡ 0:6 log m: This is a linear equation in log m: Its graph is a straight line with slope ¡0:6 and vertical intercept log k: 73. (a) The total number of individuals in the community is 50 + 50, or 100. Let P1 = 50 = 0:5; P2 = 0:5: 100 H = ¡1[P1 ln P1 + P2 ln P2 ] = ¡1[0:5 ln 0:5 + 0:5 ln 0:5] ¼ 0:693 (b) For 2 species, the maximum diversity is ln 2. (c) Yes, ln 2 ¼ 0:693. each: P1 = P2 = P3 = 1 3 H = ¡(P1 ln P1 + P2 ln P2 + P3 ln P3 ) μ ¶ 1 1 ln = ¡3 3 3 1 = ¡ ln 3 ¼ 1:099 (b) 4 species, 1 4 each: P1 = P2 = P3 = P4 = x ¼ 21:29 2009 + 21:29 = 2030:29 Therefore, on July 1, 2031, the job at Crabapple, Inc., will pay more. 1 3 1 4 H = (P1 ln P1 + P2 ln P2 + P3 ln P3 + P4 ln P4 ) μ ¶ 1 1 ln = ¡4 4 4 1 = ¡ ln 4 ¼ 1:386 (c) Notice that ¡ ln 1 = ln(3¡1 )¡1 = ln 3 ¼ 1:099 3 ¡ ln 1 = ln(4¡1 )¡1 = ln 4 ¼ 1:386 4 and by Property c of logarithms, so the populations are at a maximum index of diversity. 76. mX + N = m logb x + logb n = logb xm + logb n = logb nxm = logb y =Y Thus, Y = mX + N: Chapter 2 77. C(t) = C0 e¡kt Substituting this value into 4:3 = a(0:3)b ; When t = 0; C(t) = 2; and when t = 3; C(t) = 1: ¡k(0) 2 = C0 e C0 = 2 1 = 2e¡3k 1 = e¡3k 2 1 ¡3k = ln = ln 2¡1 = ¡ ln 2 2 ln 2 k= 3 C2 1 ln k C1 1 5 C1 T = ln 2 ln C1 3 T = T = 3 ln 5 ln 2 T ¼ 7:0 The drug should be given about every 7 hours. 78. (a) From the given graph, when x = 0:3 kg y ¼ 4:3 ml= min; and when x = 0:7 kg y ¼ 7:8 ml= min : (b) If y = axb ; then ln y = ln (axb ) = ln a + b ln x: Thus, there is a linear relationship between ln y and ln x: (c) NONLINEAR FUNCTIONS 4:3 = a(0:3)b 7:8 = a(0:7)b a(0:3)b 4:3 = 7:8 a(0:7)b μ ¶b 0:3 4:3 = 7:8 0:7 μ ¶ μ ¶b 4:3 0:3 ln = ln 7:8 0:7 μ ¶ μ ¶ 4:3 0:3 ln = b ln 7:8 0:7 ¡ ¢ ln 4:3 ¡ 7:8 ¢ b= ln 0:3 0:7 b ¼ 0:7028 4:3 = a(0:3)0:7028 a= 4:3 ¼ 10:02: (0:3)0:7028 Therefore, y = 10:02x0:7028 : (d) If x = 0:5, y = 10:02(0:5)0:7028 ¼ 6:16: We predict that the oxygen consumption for a guinea pig weighing 0.5 kg will be about 6.16 ml/min. 79. (a) h(t) = 37:79(1:021)t Double the 2005 population is 2(42.69) = 85.38 million 85:38 = 37:79(1:021)t 85:38 = (1:021)t 37:79 μ ¶ 85:38 log1:021 =t 37:79 ¢ ¡ ln 85:38 37:79 t= ln 1:021 ¼ 39:22 The Hispanic population is estimated to double their 2005 population in 2039. (b) h(t) = 11:14(1:023)t Double the 2005 population is 2(12.69) = 25.38 million 25:38 = 11:14(1:023)t 25:38 = (1:023)t 11:14 μ ¶ 25:38 log1:023 =t 11:14 ¢ ¡ ln 25:38 11:14 t= ln 1:023 ¼ 36:21 The Asian population is estimated to double their 2005 population in 2036. Section 2.5 Logarithmic Functions 80. N (r) = ¡5000 ln r (f ) I0 = 0:0002 microbars (a) N(0:9) = ¡5000 ln (0:9) ¼ 530 (b) N (0:5) = ¡5000 ln (0:5) ¼ 3500 (c) N(0:3) = ¡5000 ln (0:3) ¼ 6000 (d) N (0:7) = ¡5000 ln (0:7) ¼ 1800 (e) ¡5000 ln r = 1000 1000 ¡5000 1 ln r = ¡ 5 ln r = 1,200,000,000I0 = 1,200,000,000(0.0002) = 240,000 microbars 895,000,000,000I0 = 895,000,000,000(0.0002) = 179,000,000 microbars 83. Let I1 be the intensity of the sound whose decibel rating is 85. (a) r = e¡1=5 r ¼ 0:8 ´ ³s +1 81. C = B log2 n ´ ³ C s = log2 +1 B n s C=B 2 = +1 n s = 2C=B ¡ 1 n I 82. Decibel rating: 10 log I0 (a) Intensity, I = 115I0 μ ¶ 115I0 10 log I0 I1 = 85 I0 I1 = 8:5 log I0 10 log log I1 ¡ log I0 = 8:5 log I1 = 8:5 + log I0 Let I2 be the intensity of the sound whose decimal rating is 75. I2 = 75 I0 I2 = 7:5 log I0 10 log log I2 ¡ log I0 = 7:5 log I0 = log I2 ¡ 7:5 Substitute for I0 in the equation for log I1 : log I1 = 8:5 + log I0 = 8:5 + log I2 ¡ 7:5 = 1 + log I2 = 10 ¢ log 115 ¼ 21 log I1 ¡ log I2 = 1 (b) I = 9,500,000I0 μ ¶ 9:5 £ 106 I0 10 log = 10 log 9:5 £ 106 I0 ¼ 70 (c) I = 1,200,000,000I0 μ ¶ 1:2 £ 109 I0 10 log = 10 log 1:2 £ 109 I0 ¼ 91 (d) I = 895,000,000,000I0 μ ¶ 8:95 £ 1011 I0 10 log = 10 log 8:95 £ 1011 I0 ¼ 120 (e) I = 109,000,000,000,000I0 μ ¶ 1:09 £ 1014 I0 10 log = 10 log 1:09 £ 1014 I0 ¼ 140 log I1 =1 I2 1 I1 : This means the intenThen II12 = 10, so I2 = 10 sity of the sound that had a rating of 75 decibels 1 as intense as the sound that had a rating of is 10 85 decibels. 84. R(I) = log I I0 (a) R(1,000,000 I0 ) 1,000,000 I0 = log I0 = log 1,000,000 =6 (b) R(100,000,000 I0 ) 100,000,000 I0 = log I0 = log 100,000,000 =8 Chapter 2 I I0 I 6:7 = log I0 I 106:7 = I0 (g) Find the energy of a magnitude 6.7 earthquake. Using the formula from part f, (c) R(I) = log 6:7 = log I ¼ 5,000,000I0 I I0 I 8:1 = log I0 I 108:1 = I0 (d) R(I) = log 126,000,000I0 1985 quake = ¼ 25 1999 quake 5,000,000I0 The 1985 earthquake had an amplitude more than 25 times that of the 1999 earthquake. (f ) R(E) = E 2 log 3 E0 E = 10:05 E0 E = 1010:05 E0 E = E0 1010:05 I ¼ 126,000,000I0 (e) NONLINEAR FUNCTIONS E 2 log 3 E0 For the 1999 earthquake E 2 log 3 E0 E 10:05 = log E0 E = 1010:05 E0 For an earthquake that releases 15 times this much energy, E = E0 (15)1010:05 : μ ¶ E0 (15)1010:05 2 R(E0 (15)1010:05 ) = log 3 E0 2 = log(15 ¢ 1010:05 ) 3 ¼ 7:5 So, it’s true that a magnitude 7.5 earthquake releases 15 times more energy than one of magnitude 6.7. 85. pH = ¡ log [H+ ] (a) For pure water: 7 = ¡ log [H+ ] ¡7 = log [H+ ] 10¡7 = [H+ ] 6:7 = For acid rain: 4 = ¡ log [H+ ] ¡4 = log [H+ ] 10¡4 = [H+ ] E = 1010:05 E0 For the 1985 earthquake, E 2 log 3 E0 E 12:15 = log E0 E = 1012:15 E0 8:1 = E = 1012:15 E0 The ratio of their energies is 1012:15 E0 = 102:1 ¼ 126 1010:05 E0 The 1985 earthquake had an energy about 126 times that of the 1999 earthquake. 10¡4 = 103 = 1000 10¡7 The acid rain has a hydrogen ion concentration 1000 times greater than pure water. (b) For laundry solution: 11 = ¡ log [H+ ] 10¡11 = [H+ ] For black co¤ee: 5 = ¡ log [H+ ] 10 = [H+ ] ¡5 10¡5 = 106 = 1,000,000 10¡11 The co¤ee has a hydrogen ion concentration 1,000,000 times greater than the laundry mixture.