Chapter 10 – Algebra 2 – Selected Answers ‘Evens’ p.1 Section 10.1 (p.457) Oral Ex/ 1a. 3 b. 1/3 2a. 2 b. ½ 3a. 4 b. ¼ 4a. 2 b. ½ 5. 1/3 6. 9 7. 8 8. 1/5 9. c 10. a 11. c 12. a 13. b 14. a 15. b 16. a 17a. False b. True 18a. False b. True 19a. False b. True 20a. False b. True 21a. False b. True 22a. False b. True 23. 2 24. 3/2 25. –3 26. –4/3 27. Divide by 3; raise to the 4th power 28. Divide by 5; raise to the –2/3 power 29. Raise to the –1/2 power; add 3 30. Raise to –2 power; divide by 5 Written Exercises 4 1 2. 3 4. ½ 6. 9 8. 1/125 10. –1/8 12. 27/8 14. –27 16. 1/49 18. 1/3 20. 8 22. p 3 q 3 24. x2 y 4 3 or x 2 y 38. a2 40. 43 26. 2b2 c 1 b 1 4 or b 14 4 3 or 2b2c 43 28. q2 or p 1q 2 30. 4 32. 3 34. p 6 32 36. 4 4 8 42. x – 2x2 44a. 1/36 b. 1/108 46a. 1/72 b. 27/8 48. 3 50. 4 Section 10.2 (p.461) Oral Ex/ 1. 71.5 2. 5 3. 42 4. 6 11. –3/2 12. ¼ Written Exercises 2a. 7 3 2 b. 49 3 c. 7 6 d. 7 3 2 5. 9 6. 1 7. 7 2 8. 3 10 9. –2 10. 3/2 4. 5 6. 16 8. 1 10. 1 12. 2 14. 1/5 16. 3 2 2 3 18. 2 2 20. 3/2 22. –1/2 24. 11/4 26. –5/9 28. 3 30. 1 32. 0, 2 38. x = 0, 1 Section 10.3 (p.465) Oral Ex/ 1a. 12 b. 3 c. –15 d. 3x + 3 2a. 13 b. 16 c. –17 d. 3x + 1 3a. 18 b. 38 c. 3x2 + 6 d. 9x2 + 2 4a. 12 b. 18 c. x2 + 3 d. x2 + 2x + 3 5. x/3 6. x – 1 7. Let h 1 or h 1 ( x) represent an inverse function. Then, since h(x) does not pass the horizontal line test (is not 1-to-1), there is no inverse (function). Some texts will let h 1 represent an inverse relation which may or may not be a function. Chapter 10 – Algebra 2 – Selected Answers ‘Evens’ Section 10.3 (continued) Written Exercises 2a. 1 b. 11 c. –3 d. 2x 3 4a. 4 b. 2 2 c. 22 x d. 4x 6a. 0 b. 4 3 3 c. 2 yes no inverse (function) x 3 d. x–6 y = 3x – 6 yes f 1( x) 12x p.2 g 1( x) x3 2 g 1 ( x) 3 x 2 x 3 2 24. f(x) = x2 is not ‘1-to-1’. It doesn’t pass the horizontal line test and therefore has no inverse. g(x) = x is one-to-one. Note the Dg x x 0 Rg 1 & Rg y y 0 Dg 1 1 g ( x) 5 To find the inverse of y = x , we interchange the x & y to get: x = y and solve this second equation for ‘y’ getting: g 1 ( x) y x 2 with the restriction that the domain of the inverse function consists of only nonnegative numbers, x 0 . The graph is the right side of the parabola. Chapter 10 – Algebra 2 – Selected Answers ‘Evens’ Section 10.4 (p.470) p.3 1 1 5. log 64 = 3 6. log 27 = 3/2 Oral Ex/ 1. 25 = 32 2. 32 = 9 3. 7 2 7 4. 34 81 4 9 7. log10 0.01 = –2 8. log16 18 34 9. 2 10. 4 11. 2 12. –2 13. 3/2 14. 0 15. 16 16. 5 17. 3; 4 18. 2; 3 19. –4; –3 20. –10, 10 Written Exercises 2. 2 4. 1 6. 2/3 8. –3 10. 5/2 12. 1/6 14. 2/5 16. –3 18. –3/2 20. 216 22. 36 6 24. 3 26. 2 28. 7 30. no solution 32a. 3; 1/3 b. ½ ; 2 c. For a, b > 0, not equal to 1: 1 34a. 4x b. 16; 1/8 c. Dg positive numbers Rg 1 & Rg eals Dg 1 log a b logb a 40. 8 42. negative Section 10.5 (p.475) Oral Ex/ 1. 4log 3 M 2. 6log3 N 3. 4log3 M log3 N 4. log3 M 3log3 N 5. log3 M 6. 12 log3 M 7. 23 log3 N 8. 32 log3 N 9. log a 12 10. loga 75 11. log a 16 12. log a 81 13. log a 6 14. log a 6 15. logb 30 16. log b 15 17. logb p 2q 18. log b 19. log b rs 20. logb x y x y3 21. c + d 22. 2c + d 23. c + 2d 24. c – d Written Exercises 2. 4log 2 M 4log 2 N 4. 23 log2 M 13 log 2 N 6. 7log 2 M 7log 2 N 8. log 2 M log 2 N 10. 0.65 12. 0.48 14. 1.55 16. 2.95 18. –3.30 20. 2.20 22. log10 x y4 26. log 5 25y x 28. log 9 3 x 30. 2 32. –2 34. 54 36. 3 38. 9 40. 5/2 42a. 3 3 b. undefined c. 0 44. 25 46. 16 48. 2 34 50. 3, 6 24. log 5 M 4 N Chapter 10 – Algebra 2 – Selected Answers ‘Evens’ p.4 Section 10.6 (p.481) Oral Ex/ 1. .6075 2. 1.6075 3. 2.6075 4. –.3925 5. .9222 6. 1.9222 7. –.0778 log 7 log8 log3 8. –1.0778 9. 4.05 10. 8.34 11. 83.4 12. .834 13. 14. 15. log 2 log 9 2log8 log 7 16. log3 Written Exercises 2. 1.79 1010 4. 6460 6. 3.04 8. 55.4 10. 2.55 12. 40.3 14. .00123 16a. log1 5 b. 1.43 18a. log 2 log1.02 b. 35.0 20a. 3 2 log12 b. 1.39 22a. 2 log 21 log 4 b. –.196 24. 2/5 26. 13/9 28. 354 30. 83,900 32. 9.92 108 34. 135 36. 1.16 38. –.356 40. 1, 1.26 log c b 42. Change-of-Base Theorem - log a b log c a Proof: We’ll use: aloga b b Since log and exponent are inverse functions, f 1 f ( x) x . Begin with log c b and substitute for ‘b’ using the result above:. Now use the ‘Down in front!’ Law log x y z z log x y to move the exponent down. logc b logc aloga b log a b logc a logc b log a b logc a . Now divide by logc a to get: log c b log a b q.e.d. log c a Section 10.7 (p.486) Oral Ex/ 1. 5(1.1)t, 40(1.05)t, $60, 7% 2. 10,000(.7)t, 300(.8)t, $250, 5% 3. 1000 2 9 , 3 106 2 25 , 7500, 12 years 4. 80 12 8 , 1200 12 30 , 500, 3 t t t t Problems 2a. $107.20 b. $141.57 c. $200.42 d. $401.69 4a. $1126.83 b. $1269.73 c. $1430.77 d. $3300.39 6a. $3150.00 b. $2066.72 c. $1220.37 d. $425.52 8a. 8N0 b. 236N0 c. 4.72 1021 N 0 10. 25 g 12. 18.9 or approximately 19 years 14. B 16. 4.64 min 18. 14.9% 20. 20.5% 22. 1.24 x 107 Chapter 10 – Algebra 2 – Selected Answers ‘Evens’ p.5 Section 10.8 (p.490) Oral Ex/ 1. e1.39 = 4 2. e1.39 14 3. e1 = e 4. ln 7.39 2 5. ln.14 2 6. ln1.22 15 1 1 7. –1 8. 12 9. ½ 10. e5 11. e 3 12. ln 3 13. ¾ 14. 10 15. e 2 16. 2.718 Written Exercises 2. e4.61 = 100 4. e2 12 6. ln 1097 = 7 8. ln 1.40 = 1/3 10. 10 12. – ½ 14. undefined e 16. 0.5 18. ln 4 20. ln 21 22. ln 16 24. e 2 26. e 28. e6 30. ln 13 32. ln 2 34. ln 17 3e 36. 12 ln5 38. ln 4 40. e, e 1 42. 1 5 44. D : x 0, R eals 2 50a. e5 b. e2 c. e 2 46. D : x 0, R eals Chapter Review (p.496) 2. c 4. d 6. b 8. c 10. d 12. a 14. b 16. c Chapter Test (p.497) 1a. 8 b. 4/3 2a. y 5 b. 125 3. 5 4a. 15 b. 4x2 – 4x + 5 5. f ( g ( x)) 3 x3 1 1 3 x3 x and g ( f ( x)) 6a. log2 8 = 3 b. 16 10. 32 3 3 x 1 1 x 1 1 x 1 7a. 3 b. 3 2 8a. 4 b. 8 9a. .903 b. 1.079 c. 1.176 64 log 2 11. 7.9 h 12. e 4 13. ln 3log5 e 7 Mixed Problem Solving (p.498) 1 2. The equation 1 x has no real solution. 4. 120 rev/min 6. 10m x 20m x 8. 2 eagles, 2 quarter-eagles; 1 eagle, 6 quarter-eagles; 10 quarter-eagles 10. 7, 8, 9, –6, –5, –4 12. 5.4 years Preparing for College Entrance Exams 2. D 4. B 6. E 8. C