Answer
$-\frac{1}{3}$
Work Step by Step
We know that if $b\gt0$ and $b\ne1$, then $y=log_{b}x$ is equivalent to $x=b^{y}$ (where $x\gt0$ and $y$ is a real number).
Therefore, $log_{8}\frac{1}{2}=-\frac{1}{3}$, because $8^{-\frac{1}{3}}=\frac{1}{8^{\frac{1}{3}}}=\frac{1}{\sqrt[3] 8}=\frac{1}{2}$.
