Answer
$log_{a}\frac{pr^{2}}{q}$
Work Step by Step
We know that $log_{b}x^{r}=rlog_{b}x$ (where $x$ and $b$ are positive real numbers, $b\ne1$, and $r$ is a real number).
Therefore, $(log_{a}p-log_{a}q)+2log_{a}r=(log_{a}p-log_{a}q)+log_{a}r^{2}$.
We know that $log_{b}\frac{x}{y}=log_{b}x-log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $(log_{a}p-log_{a}q)+log_{a}r^{2}=log_{a}\frac{p}{q}+log_{a}r^{2}$.
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, $log_{a}\frac{p}{q}+log_{a}r^{2}=log_{a}\frac{p\times r^{2}}{q}=log_{a}\frac{pr^{2}}{q}$.
