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: 83



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 can re-write the expression as: $3^{2x}=(3^{-x})^{-2}$ We will use Rule-(b) as: $3^{-x(-2)}=(3^{-x})^{-2}$ Since, $3^{-x}=2$, then we simplify the expression as: $(3^{-x})^{-2}=2^{-2}$ Now, we will use Rule-(a) as: $2^{-2}=\dfrac{1}{2^2}=\dfrac{1}{4}$ Therefore, $3^{2x}=\dfrac{1}{4}$
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