## Algebra: A Combined Approach (4th Edition)

We are given $log_{b}3=.5$ and $log_{b}5=.7$. 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, $log_{b}15=log_{b}(5\times3)=log_{b}5+log_{b}3=.7+.5=1.2$.