Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.3 Exponential Functions - 4.3 Assess Your Understanding - Page 307: 82



Work Step by Step

In order to solve the given problem, we will use the following two rules: $ (a) a^{-p}=\dfrac{1}{a^p} \\ (b) a^{pq}=(a^p)^q$ We will use Rule-(b) as: $4^{-x}=(2^2)^{-x}=2^{-2x}$ Now, we will use Rule-(a) as: $2^{-2x}=\dfrac{1}{2^{2x}}$ So, $4^{-x}=\dfrac{1}{(2^x)^2}=(2^x)^{-2}$ Since, $2^x=3$, then we simplify the expression as: $4^{-x}=3^{-2}$ Therefore, $4^{-x}=\dfrac{1}{3^2} \\ 4^{-x}=\dfrac{1}{9}$
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