#### Answer

1.5562

#### 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, $log_{10}36=log_{10}(9\times4)=log_{10}9+log_{10}4$.
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, $log_{10}9+log_{10}4=log_{10}9+log_{10}2^{2}=log_{10}9+2log_{10}2=.9542+2(.3010)=.9542+.6020=1.5562$.