chapter8ProblemsAndSolutions-1

chapter8ProblemsAndSolutions-1 - Chapter 8 Exponential...

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Chapter 8 Exponential functions 8.1 Graph the following functions: (a) f ( x ) = x 2 e - x (b) f ( x ) = ln( x 2 + 3) (c) f ( x ) = ln( e 2 x ) Detailed Solution: See Figures 8.1. (a) x y (b) x y (c) y x Figure 8.1: Figures for solution to Problem 8.1 8.2 Express the following in terms of base e : (a) y = 3 x (b) y = 1 7 x (c) y = 15 x 2 +2 Express the following in terms of base 2: v.2005.1 - September 4, 2009 1
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Math 102 Problems Chapter 8 (d) y = 9 x (e) y = 8 x (f) y = - e x 2 +3 Express the following in terms of base 10: (g) y = 21 x (h) y = 1000 - 10 x (i) y = 50 x 2 - 1 Detailed Solution: (a) y = e ln(3 x ) = e x ln(3) (b y = 7 - x = e ln(7 - x ) = e - x ln(7) (c) y = e ln(15 x 2 +2 ) = e ( x 2 +2) ln(15) (d) y = 2 log 2 (9 x ) = 2 x log 2 (9) (e) y = 2 log 2 (8 x ) = 2 x log 2 (8) = 2 3 x (f) y = - 2 log 2 ( e x 2 +3 ) = - 2 ( x 2 +3) log 2 ( e ) (g) y = 10 log 10 (21 x ) = 10 x log 10 (21) (h) y = 10 log 10 (1000 - 10 x ) = 10 - 10 x log 10 (1000) = 10 - 10 x (3) = 10 - 30 x (i) y = 10 log 10 (50 x 2 - 1 ) = 10 ( x 2 - 1) log 10 (50) 8.3 Compare the values of each pair of numbers (i.e. indicate which is larger): (a) 5 0 . 75 , 5 0 . 65 (b) 0 . 4 - 0 . 2 , 0 . 4 0 . 2 (c) 1 . 001 2 , 1 . 001 3 (d) 0 . 999 1 . 5 , 0 . 999 2 . 3 v.2005.1 - September 4, 2009 2
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Math 102 Problems Chapter 8 Detailed Solution: The exponential function y = a x increases on ( -∞ , ) when the base a > 1 and decreases on ( -∞ , ) when the base 0 < a < 1. (a) Base 5 is greater than 1, so y = 5 x increases as x increases. The exponents 0 . 75 > 0 . 65, so 5 0 . 75 > 5 0 . 65 . (b) Base 0 . 4 is between 0 and 1, so y = 0 . 4 x decreases as x increases. The exponents - 0 . 2 < 0 . 2, so 0 . 4 - 0 . 2 > 0 . 4 0 . 2 . (c) Base 1 . 001 is greater than 1, so y = 1 . 001 x increases as x increases. The exponents 2 < 3, so 1 . 001 2 < 1 . 001 3 . (d) Base 0 . 999 is between 0 and 1, so y = 0 . 999 x decreases as x increases. The exponents 1 . 5 < 2 . 3, so 0 . 999 1 . 5 > 0 . 999 2 . 3 . 8.4 Rewrite each of the following equations in logarithmic form: (a) 3 4 = 81 (b) 3 - 2 = 1 9 (c) 27 - 1 3 = 1 3 Detailed Solution: (a) log 3 81 = 4 (b) log 3 1 9 = - 2 (c) log 27 1 3 = - 1 3 8.5 Solve the following equations for x : (a) ln x = 2 ln a + 3 ln b (b) log a x = log a b - 2 3 log a c v.2005.1 - September 4, 2009 3
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Math 102 Problems Chapter 8 Detailed Solution: (a) ln x = ln a 2 + ln b 3 = ln a 2 b 3 x = a 2 b 3 (b) log a x = log a b - log a c 2 3 = log a b c 2 3 x = b c 2 3 8.6 Reflections and transformations What is the relationship between the graph of y = 3 x and the graph of each of the following functions? (a) y = - 3 x (b) y = 3 - x (c) y = 3 1 - x (d) y = 3 | x | (e) y = 2 · 3 x (e) y = log 3 x Detailed Solution: (a) reflected about the x -axis (b) reflected about the y -axis (c) reflected about the y -axis and shifted along the positive x -axis by 1 unit (d) y = braceleftbigg 3 x , x 0 3 - x , x < 0 The graph of y = 3 | x | is the same as y = 3 x for x 0, and is the same as y = 3 - x for x < 0. (e) stretched along the y -axis by a factor of 2 (f) y = log 3 x is the inverse function of y = 3 x , so its graph is the reflection of the graph of y = 3 x about the line y = x . 8.7 Solve the following equations for x : (a) e 3 - 2 x = 5 (b) ln(3 x - 1) = 4 (c) ln(ln( x )) = 2 (d) e ax = Ce bx , where a negationslash = b and C > 0. v.2005.1 - September 4, 2009 4
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Math 102 Problems Chapter 8 Detailed Solution: (a) 3 - 2 x = ln(5), hence x = 3 - ln(5) 2 (b) 3 x - 1 = e 4 , hence x = e 4 +1 3 (c) ln( x ) = e 2 , hence x = e ( e 2 ) = e e · e (and not e 2 e !) (d) ax = ln( Ce bx ) = ln( C ) + bx , hence x = ln( C ) a - b 8.8 Find the first derivative for each of the following functions: (a) y = ln(2 x + 3) 3 (b) y = ln 3 (2 x + 3) (c) y = ln(cos 1 2 x ) (d) y = log a ( x 3 - 2 x ) (Hint : d dx (log a x ) = 1 x ln a ) (e) y = e 3 x 2 (f) y = a - 1 2 x
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