## College Algebra (6th Edition)

$C+A$
We are given that $log_{b}2=A$ and $log_{b}3=C$. Based on the product rule of logarithms, we know that $log_{b}(MN)=log_{b}M+log_{b}N$ (for $M\gt0$ and $N\gt0$). Therefore, $log_{b}6=log_{b}(3\times2)=log_{b}3+log_{b}2=C+A$.