## Precalculus (6th Edition) Blitzer

$\dfrac{1}{2}$
RECALL: $\log_b{b^x} = x.$ Note that $9=\sqrt{81} = 81^{\frac{1}{2}}.$ Thus, the given expression can be written as $\log_{81}{81^{\frac{1}{2}}}.$ Use the rule above to obtain: $\dfrac{1}{2}.$