Answer
2.2094
Work Step by Step
We are given that $log_{10}2=.3010$ and $log_{10}9=.9542$.
We know that $log_{b}xy=log_{b}x+log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $log_{10}162=log_{10}(81\times2)=log_{10}81+log_{10}2$.
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, $log_{10}81+log_{10}2=log_{10}9^{2}+log_{10}2=2log_{10}9+log_{10}2=2(.9542)+.3010=1.9084+.3010=2.2094$.
