Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Section 12.5 - Logarithmic Functions - Exercise Set - Page 875: 68



Work Step by Step

We know that if $b\gt0$ and $b\ne1$, then $y=log_{b}x$ is equivalent to $x=b^{y}$ (where $x\gt0$ and $y$ is a real number). Therefore, $log_{x}2=-\frac{1}{3}$ is equivalent to $x^{-\frac{1}{3}}=2$. Therefore, $x=\frac{1}{8}$, because $(\frac{1}{8})^{-\frac{1}{3}}=\frac{1}{(\frac{1}{8})^{\frac{1}{3}}}=\frac{1}{\sqrt[3]\frac{1}{8}}=\frac{1}{\frac{1}{2}}=2$.
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