# Unit1 - Unit 1: Exponential Functions and Logarithmic...

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Unit 1: Exponential Functions and Logarithmic Functions Defnition 1.1. An exponential function is a function of the form f ( x ) = b x where b is a positive constant called the base of the function. For instance, f ( x ) = 2 x is an exponential function, with base 2. Notice: For any positive base b , the exponential function f ( x ) = b x has do- main D f = ( −∞ , ), and has range R f = (0 , ) provided that b n = 1. (For b = 1, so that f ( x ) = b x = 1 x , we have f ( x ) = 1 for all values of x .) Theorem 1.2. The following are properties of exponents: Properties of Exponents: (i) b x b y = b x + y (ii) b x /b y = b x y (iii) ( b x ) y = b xy Example 1 . Simplify the following: (a) p 3 5 3 7 27 P 2 (b) p 7 6 7 4 7 2 P 3 1

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Solution: (a) p 3 5 3 7 27 P 2 = p 3 (5+7) 3 3 P 2 (b) p 7 6 7 4 7 2 P 3 = p 7 6+( 4) 7 2 P 3 = ( 3 (5+7) 3 ) 2 = ( 7 (6 4) 2 ) 3 = 3 (12 3) × 2 = 7 (2 2) × 3 = 3 (9 × 2) = 7 0 = 3 18 = 1 Another class of function, the logarithmic function , is closely related to the exponential function. Defnition 1.3. For any b > 0, b n = 1, we de±ne the logarithm of x to base b , denoted log b x , as the exponent to which b must be raised in order to get the value x . That is, b log b x = x or y = log b x ⇐⇒ b y = x Notice: The function y = log b x is the inverse of the function y = b x , so the function y = log b x has domain (0 , ) and range ( −∞ , ). Example 2 . Evaluate the following logarithmic expressions. (a) log 2 (16) (b) log 2 ( 1 8 ) (c) log 3 (9) (d) log b (1) (e) log 4 (8) Solution: (a) Since 16 = 2 4 , then log 2 (16) = log 2 (2 4 ) = 4. That is, the power to which 2 must be raised to get 16 is 4. (b) 1 8 = 1 2 3 = 2 3 , so log 2 ( 1 8 ) = 3. (c) 9 = 3 2 , so log 3 (9) = 2. (d) 1 = b 0 for any b > 0, so log b (1) = log b ( b 0 ) = 0 for any b > 0. (e) 8 = 2 3 and 2 = 4 = 4 1 2 , so then 8 = (4 1 2 ) 3 = 4 3 2 . Therefore, log 4 (8) = 3 2 . 2
Theorem 1.4. The following are properties of logarithms: Properties of Logarithms: (i) log b ( uv ) = log b ( u ) + log b ( v ) (ii) log b ( u/v ) = log b ( u ) log b ( v ) (iii) log b ( u r ) = r log b ( u ) All of these properties follow directly from the properties of exponents. They can easily be proven by converting from logarithms to exponents. For instance, for the ±rst property, let x = log b u . Then u = b x . Likewise, let y = log b v so that v = b y . Then log b ( uv ) = log b ( b x b y ) = log b ( b x + y ) = x + y = log b u + log b v . Example 3 . Evaluate log 2 6 log 2 75 + 2 log 2 5 Solution: log 2 6 log 2 75 + 2 log 2 5 = (log 2 6 log 2 75) + log 2 (5 2 ) = log 2 ( 6 75 ) + log 2 25 = log 2 b( 2 25 ) × 25 B = log 2 2 = 1 Example 4 . If log 2 (2 x 2) = 4, ±nd x .

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## This note was uploaded on 11/03/2009 for the course MATH 0110A taught by Professor Olds,vicky during the Fall '09 term at UWO.

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Unit1 - Unit 1: Exponential Functions and Logarithmic...

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