Precalculus (6th Edition) Blitzer

$-3$
RECALL: $\log_b{b^x} = x.$ Note that $\dfrac{1}{8}=\dfrac{1}{2^3} = 2^{-3}.$ Thus, the given expression can be written as $\log_2{2^{-3}}.$ Use the rule above t obtain $-3.$