Intermediate Algebra (12th Edition)

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, $log_{10}9^{5}=5log_{10}9=5(.9542)=4.771$.