# 5_4 - Chapter 5 Exponential and Logarithmic Functions 5.4 1...

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Chapter 5 678 Exponential and Logarithmic Functions 5.4 Logarithmic Functions 1. 3 x - 7 8 - 2 x 5 x 15 x 3 The solution set is x x 3 { } . 2. x 2 - x - 6 0 x - 3 ( 29 x + 2 ( 29 0 f ( x ) = x 2 - x - 6 x = - 2 and x = 3 are the zeros of f . Interval -∞ , - 2 ( 29 - 2,3 ( 29 3, ( 29 Number Chosen –3 0 4 Value of f 6 –6 6 Conclusion Positive Negative Positive The solution set is x x < - 2 or x 3 { } . 3. x - 1 x + 4 0 f x ( 29 = x - 1 x + 4 f is zero or undefined when x = 1 or x = - 4 . Interval -∞ , - 4 ( 29 - 4,1 ( 29 1, ( 29 Number Chosen –5 0 2 Value of f 6 –0.25 1 6 Conclusion Positive Negative Positive The solution set is x x < - 4 or x 1 { } . 4. x x 0 { } 5. 1,0 ( 29 , a ,1 ( 29 , 1 a , - 1 6. 1 7. False 8. True 9. 9 = 3 2 is equivalent to 2 = log 3 9 10. 16 = 4 2 is equivalent to 2 = log 4 16 11. a 2 = 1.6 is equivalent to 2 = log a 1.6 12. a 3 = 2.1 is equivalent to 3 = log a 2.1 13. 1.1 2 = M is equivalent to 2 = log 1.1 M 14. 2.2 3 = N is equivalent to 3 = log 2.2 N

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Section 5.4 Logarithmic Functions 679 15. 2 x = 7.2 is equivalent to x = log 2 7.2 16. 3 x = 4.6 is equivalent to x = log 3 4.6 17. x 2 = π is equivalent to 2 = log x π 18. x π = e is equivalent to π = log x e 19. e x = 8 is equivalent to x = ln8 20. e 2.2 = M is equivalent to 2.2 = ln M 21. log 2 8 = 3 is equivalent to 2 3 = 8 22. log 3 1 9  = - 2 is equivalent to 3 - 2 = 1 9 23. log a 3 = 6 is equivalent to a 6 = 3 24. log b 4 = 2 is equivalent to b 2 = 4 25. log 3 2 = x is equivalent to 3 x = 2 26. log 2 6 = x is equivalent to 2 x = 6 27. log 2 M = 1.3 is equivalent to 2 1.3 = M 28. log 3 N = 2.1 is equivalent to 3 2.1 = N 29. log 2 π = x is equivalent to 2 ( 29 x = π 30. log π x = 1 2 is equivalent to π 1/2 = x 31. ln4 = x is equivalent to e x = 4 32. ln x = 4 is equivalent to e 4 = x 33. log 2 1 = 0 since 2 0 = 1 34. log 8 8 = 1 since 8 1 = 8 35. log 5 25 = 2 since 5 2 = 25 36. log 3 1 9  = - 2 since 3 - 2 = 1 9 37. log 1/2 16 = - 4 since 1 2 - 4 = 2 4 = 16 38. log 1/3 9 = - 2 since 1 3 - 2 = 3 2 = 9 39. log 10 10 = 1 2 since 10 = 10 40. log 5 25 3 = 2 3 since 5 2/3 = 25 = 25 3 41. log 2 4 = 4 since 2 ( 29 4 = 4 42. log 3 9 = 4 since 3 ( 29 4 = 9 43. ln e = 1 2 since e = e 44. ln e 3 = 3 since e 3 = e 3 45. f ( x ) = ln( x - 3) requires x - 3 0 x 3 The domain of f is x x 3 { } . 46. g ( x ) = ln( x - 1) requires x - 1 0 x 1 The domain of g is x x 1 { } .
Chapter 5 Exponential and Logarithmic Functions 680 47. F ( x ) = log 2 x 2 requires x 2 0 . x 2 0 for all x 0. The domain of F is x x 0 { } . 48. H ( x ) = log 5 x 3 requires x 3 x 3 0 for all x The domain of H is x x 0 { } . 49. f x ( 29 = 3 - 2log 4 x 2 requires x 2 x 2 0 x 0 The domain of f is x x 0 { } . 50. g x ( 29 = 8 + 5ln 2 x ( 29 requires 2 x 2 x 0 x 0 The domain of g is x x 0 { } . 51. f ( x ) = ln 1 x + 1 requires 1 x + 1 0 p x ( 29 = 1 x + 1 p is undefined when x = - 1. Interval -∞ , - 1 ( 29 - 1, ( 29 Number Chosen –2 0 Value of p –1 1 Conclusion Negative Positive The domain of f is x x - 1 { } . 52. g ( x ) = ln 1 x - 5 requires 1 x - 5 0 p x ( 29 = 1 x - 5 p is undefined when x = 5. Interval -∞ ,5 ( 29 5, ( 29 Number Chosen 4 6 Value of p –1 1 Conclusion Negative Positive The domain of g is x x 5 { } .

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Section 5.4 Logarithmic Functions 681 53.
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## This note was uploaded on 04/07/2008 for the course MAC 1105 taught by Professor Any during the Spring '08 term at FIU.

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5_4 - Chapter 5 Exponential and Logarithmic Functions 5.4 1...

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