Answer
-2
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, $log_{3}\frac{1}{9}=-2$, because $3^{-2}=\frac{1}{3^{2}}=\frac{1}{3\times3}=\frac{1}{9}$.