Answer
-3
Work Step by Step
From the definition of the logarithmic function on page 456, we know that $y=log_{b}x$ is equivalent to $b^{y}=x$(for $x\gt0$, $b\gt0$, and $b\ne1$).
Therefore, $log_{2}\frac{1}{8}=-3$, because $2^{-3}=\frac{1}{2^{3}}=\frac{1}{2\times2\times2}=\frac{1}{8}$.