Answer
$-\displaystyle \frac{1}{12}$
Work Step by Step
From the definition of logarithms,
for $a>0$, $a\neq 1$, and $x > 0$,
$y=\log_{a}x$ means $a^{y}=x$.
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$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 $2=8^{1/3},$
$=(8^{1/3})^{-1/4}=8^{-1/12},$
$y=-\displaystyle \frac{1}{12}$
