Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.3 - Logarithmic Functions - 9.3 Exercises: 81

Answer

$\frac{1}{3}$

Work Step by Step

For any positive real number $b$ (such that $b\ne1$), we know that $log_{b}b^{r}=r$ (as long as $r\gt0$). For $log_{6}\sqrt[3] 6$: We know that $6^{\frac{1}{3}}=\sqrt[3] 6$, so we can replace $\sqrt[3] 6$ with $6^{\frac{1}{3}}$. Therefore, $log_{6}\sqrt[3] 6=log_{6}6^{\frac{1}{3}}=\frac{1}{3}$.
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