43 let p x0 y0 be a point where a line through the

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Unformatted text preview: ) = 4x3 + 3x2 . Thus F (3) = 2f (2)(1/7) = 2(44)(1/7) = 88/7. EXERCISE SET 4.2 1. (a) −4 (b) 4 (c) 1/4 2. (a) 1/16 (b) 8 (c) 1/3 3. (a) 2.9690 (b) 0.0341 4. (a) 1.8882 (b) 0.9381 5. (a) log2 16 = log2 (24 ) = 4 (b) (c) log4 4 = 1 (d) 1 = log2 (2−5 ) = −5 32 log9 3 = log9 (91/2 ) = 1/2 log2 6. (a) log10 (0.001) = log10 (10−3 ) = −3 (c) ln(e3 ) = 3 (b) log10 (104 ) = 4 √ (d) ln( e) = ln(e1/2 ) = 1/2 7. (a) 1.3655 (b) −0.3011 8. (a) −0.5229 (b) 9. (a) 2 ln a + 10. (a) 1 ln c − ln a − ln b = t/3 − r − s 3 11. (a) 1 + log x + 12. (a) 13. log 1 1 ln b + ln c = 2r + s/2 + t/2 2 2 1.1447 (b) ln b − 3 ln a − ln c = s − 3r − t (b) 1 (ln a + 3 ln b − 2 ln c) = r/2 + 3s/2 − t 2 1 log(x − 3) 2 (b) 2 ln |x| + 3 ln sin x − 1 log(x + 2) − log cos 5x 3 (b) 1 1 ln(x2 + 1) − ln(x3 + 5) 2 2 24 (16) = log(256/3) 3 14. log 1 ln(x2 + 1) 2 √ √ 100 x x − log(sin3 2x) + log 100 = log sin3 2x Exercise Set 4.2 √ 3 15. 17. ln √ 104 x(x + 1)2 cos x 16. 1 + x = 103 = 1000, x = 999 x = 10−1 = 0.1, x = 0.01 18. x2 = e4 , x = ±e2 19. 1/x = e−2 , x = e2 20. x=7 21. 2x = 8, x = 4 22. log10 x3 = 30, x3 = 1030 , x = 1010 23. log10 x = 5, x = 105 24. ln 4x − ln x6 = ln 2, ln 25. ln 2x2 = ln 3, 2x2 = 3, x2 = 3/2, x = equation) 26. ln 3x = ln 2, x ln 3 = ln 2, x = 27. ln 5−2x = ln 3, −2x ln 5 = ln 3, x = − 28. 1 e−2x = 5/3, −2x = ln(5/3), x = − ln(5/3) 2 29. e3x = 7/2, 3x = ln(7/2), x = 30. ex (1 − 2x) = 0 so ex = 0 (impossible) or 1 − 2x = 0, x = 1/2 31. e−x (x + 2) = 0 so e−x = 0 (impossible) or x + 2 = 0, x = −2 32. e2x − ex − 6 = (ex − 3)(ex + 2) = 0 so ex = −2 (impossible) or ex = 3, x = ln 3 33. e−2x − 3e−x + 2 = (e−x − 2)(e−x − 1) = 0 so e−x = 2, x = − ln 2 or e−x = 1, x = 0 34. (a) √ 4 4 = ln 2, 5 = 2, x5 = 2, x = 5 2 5 x x 3/2 (we discard − 3/2 because it does not satisfy the original ln 2 ln 3 ln 3 2 ln 5 1 ln(7/2) 3 (b) y y 2 2 x 6 2 -2 -2 35. (a) 2 (b) y x y 2 6 4 x -4 2 2 -2 2 4 x -4 105 36. Chapter 4 (a) (b) y -1 y x x 3 -1 -10 37. 38. log2 7.35 = (log 7.35)/(log 2) = (ln 7.35)/(ln 2) ≈ 2.8777; log5 0.6 = (log 0.6)/(log 5) = (ln 0.6)/(ln 5) ≈ −0.3174 39. 10 2 0 0 -5 3 2 -3 41. (a) Let X = logb x and Y = loga x. Then bX = x and aY = x so aY = bX , or aY /X = b, which means loga x loga x = loga b, logb x = . loga b = Y /X . Substituting for Y and X yields logb x loga b (b) 40. Let x = a to get logb a = (loga a)/(loga b) = 1/(loga b) so (loga b)(logb a) = 1. (log2 81)(log3 32) = (log2 [34 ])(log3 [25 ]) = (4 log2 3)(5 log3 2) = 20(log2 3)(log3 2) = 20 x = 3.6541, y = 1.2958 2 2 0.6 6 42. Since the units are billions, one trillion is 1,000 units. Solve 1000 = 0.051517(1.1306727)x for x by taking common logarithms, resulting in 3 = log 0.051517 + x log 1.1306727, which yields x ≈ 77.4, so the debt first reached one trillion dollars around 1977. 43. (a) no, the curve passes through the origin (c) y = 2−x (b) y = 2x/4 √ (d) y = ( 5)x 5 -1 2 0 Exercise Set 4.2 44. (a) 106 As x → +∞ the function grows very slowly, but it is always increasing and tends to +∞. As x → 1+ the function tends to −∞. (b) y x 150 -2 45. log(1/2) < 0 so 3 log(1/2) < 2 log(1/2) 46. Let x = logb a and y = logb c, so a = bx and c = by . First, ac = bx by = bx+y or equivalently, logb (ac) = x + y = logb a + logb c. Secondly, a/c = bx /by = bx−y or equivalently, logb (a/c) = x − y = logb a − logb c. Next, ar = (bx )r = brx or equivalently, logb ar = rx = r logb a. Finally, 1/c = 1/by = b−y or equivalently, logb (1/c) = −y = − logb c. 47. 75e−t/125 = 15, t = −125 ln(1/5) = 125 ln 5 ≈ 201 days. 48. (a) If t = 0, then Q = 12 grams (c) 12e−0.055t = 6, e−0.055t = 0.5, t = −(ln 0.5)/(0.055) ≈ 12.6 hours 49. (a) 7.4; basic 50. (a) log[H + ] = −2.44, [H + ] = 10−2.44 ≈ 3.6 × 10−3 mol/L (b) log[H + ] = −8.06, [H + ] = 10−8.06 ≈ 8.7 × 10−9 mol/L (b) (b) 4.2; acidic Q = 12e−0.055(4) = 12e−0.22 ≈ 9.63 grams (c) 6.4; acidic 51. (a) 140 dB; damage (c) 80 dB; no damage 52. Suppose that I1 = 3I2 and β1 = 10 log10 I1 /I0 , β2 = 10 log10 I2 /I0 . Then I1 /I0 = 3I2 /I0 , log10 I1 /I0 = log10 3I2 /I0 = log10 3 + log10 I2 /I0 , β1 = 10 log10 3 + β2 , β1 − β2 = 10 log10 3 ≈ 4.8 decibels. 53. (d) 5.9; acidic (b) 120 dB; damage (d) 75 dB; no damage Let IA and IB be the intensities of the automobile and blender, respectively. Then log10 IA /I0 = 7 and log10 IB /I0 = 9.3, IA = 107 I0 and IB = 109.3 I0 , so IB /IA = 102.3 ≈ 200. 54. The decibel level of the nth echo is 120(2/3)n ; log 12 log(1/12) = ≈ 6.13 so 6 echoes can be heard. 120(2/3)n < 10 if (2/3)n < 1/12, n > log(2/3) log 1.5 55. (a) log E = 4.4 + 1.5(8.2) = 16.7, E = 1016.7 ≈ 5 × 1016 J (b) Let M1 and M2 be the magnitudes of earthquakes with energies of E and 10E , respectively. Then 1.5(M2 − M1 ) = log(10E ) − log E = log 10 = 1, M2 − M1 = 1/1.5 = 2/3 ≈ 0.67. 56. Let E1 and E2 be the energies of earthquakes with magnitudes M and M + 1, respectively. Then log E2 − log E1 = log(E2 /E1 ) = 1.5, E2 /E1 = 101.5 ≈ 31.6. 57. If t = −2x, then x...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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