## Calculus with Applications (10th Edition)

$-\displaystyle \frac{1}{12}$
From the definition of logarithms, for $a>0$, $a\neq 1$, and $x > 0$, $y=\log_{a}x$ means $a^{y}=x$. --------- $y=\displaystyle \log_{8}(\sqrt[4]{\frac{1}{2}})\ \$ means $\ \ 8^{y}=\displaystyle \sqrt[4]{\frac{1}{2}} .$ Since $\displaystyle \sqrt[4]{\frac{1}{2}}=(\frac{1}{2})^{1/4}=(2^{-1})^{1/4}$ ... now, since $2^{3}=8$, it follows that $2=8^{1/3},$ $=(8^{1/3})^{-1/4}=8^{-1/12},$ $y=-\displaystyle \frac{1}{12}$