College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Summary, Review, and Test - Review Exercises - Page 510: 20

Answer

-2

Work Step by Step

We know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (where $x\gt0$, $b\gt0$, and $b\ne1$). Therefore, $log_{5}\frac{1}{25}=-2$, because $5^{-2}=\frac{1}{5^{2}}=\frac{1}{5\times5}=\frac{1}{25}$.
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