## Intermediate Algebra (12th Edition)

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, $log_{10}18=log_{10}(9\times2)=log_{10}9+log_{10}2=.3010+.9542=1.2552$.