#### Answer

-.9542

#### Work Step by Step

We are given that $log_{10}2=.3010$ and $log_{10}9=.9542$.
We know that $log_{b}x^{r}=rlog_{b}x$ (where $x$ and $b$ are positive real numbers, $b\ne1$, and $r$ is a real number).
Therefore, $log_{10}\frac{1}{9}=log_{10}9^{-1}=-log_{10}9=-(.9542)=-.9542$.