## Intermediate Algebra (12th Edition)

$log_{10}7+log_{10}8$
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, $log_{10}(7\times8)=log_{10}7+log_{10}8$.