Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - Test: Chapter 12 - Page 845: 26

Answer

$x=-5$

Work Step by Step

The given expression, $ 2^x=\dfrac{1}{32} ,$ is equivalent to \begin{align*} 2^x&=\dfrac{1}{2^5} \\\\ 2^x&=2^{-5} &\left(\dfrac{1}{x^{a}}=x^{-a}\right) .\end{align*} With both sides of the equation expressed in the same base, the exponents can be equated. That is, \begin{align*} x=-5 .\end{align*}
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