Answer
False
Work Step by Step
$\displaystyle \log_{a}(\frac{P}{Q})^{x}=\qquad$ ... apply $\log_{a}M^{p}=p\cdot\log_{a}M$
$=x\displaystyle \log_{a}(\frac{P}{Q})\qquad$ ... apply $\displaystyle \log_{a}\frac{M}{N}=\log_{a}M-\log_{a}N$
$=x[\log_{a}P-\log_{a}Q]$
$=x\log_{a}P-\fbox{$x$}\log_{a}Q$