Answer
$\log_{2}\dfrac{1}{32}=-5$
Work Step by Step
Since $b^y=x$ is equivalent to $y=\log_bx,$ the given equation, $
2^{-5}=\dfrac{1}{32}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\log_{2}\dfrac{1}{32}=-5
.\end{array}
Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
by Bittinger, Marvin L.; Ellenbogen, David J.; Johnson, Barbara L.
Published by
Pearson
ISBN 10:
0-32184-874-8
ISBN 13:
978-0-32184-874-1
Chapter 12 - Exponential Functions and Logarithmic Functions - 12.3 Logarithmic Functions - 12.3 Exercise Set - Page 803: 68
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