Answer
$2log_{b}x+log_{b}y-2log_{b}z$
Work Step by Step
Based on the quotient rule of logarithms, we know that $log_{b}(\frac{M}{N})=log_{b}M-log_{b}N$ (where $b$, $M$, and $N$ are positive real numbers and $b\ne1$).
Therefore, $log_{b}(\frac{x^{2}y}{z^{2}})=log_{b}x^{2}y-log_{b}z^{2}$.
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, $log_{b}x^{2}y-log_{b}z^{2}=log_{b}x^{2}+log_{b}y-log_{b}z^{2}$.
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, $log_{b}x^{2}+log_{b}y-log_{b}z^{2}=2log_{b}x+log_{b}y-2log_{b}z$.