## Intermediate Algebra (12th Edition)

$\frac{1}{3}$
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, $log_{6}\sqrt[3] 6=log_{6}6^{\frac{1}{3}}=\frac{1}{3}$.