Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 9 - Section 9.5 - Logarithmic Functions - Exercise Set - Page 572: 25

Answer

$log_{4}\frac{1}{16}=-2$

Work Step by Step

We are given $10^{-2}=\frac{1}{100}$. If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x$ and every real number $y$. Therefore, $4^{-2}=\frac{1}{4^{2}}=\frac{1}{16}$ is equivalent to $log_{4}\frac{1}{16}=-2$.
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