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: 34

Answer

$-\frac{1}{3}$

Work Step by Step

If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x$ and every real number $y$. Therefore, $log_{8}\frac{1}{2}=-\frac{1}{3}$. We know this because $8^{-\frac{1}{3}}=\frac{1}{8^{\frac{1}{3}}}=\frac{1}{\sqrt[3] 8}=\frac{1}{2}$ (where $\sqrt[3] 8=2$, because $2^{3}=8$).
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