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: 68



Work Step by Step

If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x\gt0$ and every real number $y$. Therefore, $log_{x}2=-\frac{1}{3}$ is equivalent to $x^{-\frac{1}{3}}=2$. $x^{-\frac{1}{3}}=\frac{1}{x^{\frac{1}{3}}}=\frac{1}{\sqrt[3] x}=2$ For $\frac{1}{\sqrt[3] x}=2$, multiply both sides by $\sqrt[3] x$. $2\sqrt[3] x=1$ Divide both sides by 2. $\sqrt[3] x=\frac{1}{2}$ $x=\frac{1}{8}$, because $(\frac{1}{2})^{3}=\frac{1}{8}$
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