Answer
$2log_{b}x+log_{b}y$
Work Step by Step
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, $log_{b}(x^{2}y)=log_{b}x^{2}+log_{b}y$.
According to the power rule of logarithms, we know that $log_{b}M^{p}=plog_{b}M$ (when $b$ and $M$ are positive real numbers, $b\ne1$, and $p$ is any real number).
Therefore, $log_{b}x^{2}+log_{b}y=2log_{b}x+log_{b}y$.