Example 2 If $00, a \neq 1$, compare $p=\left|\log _{a}(1-x)\right|$ and $q=\left|\log _{a}(1+x)\right|$.
Solve $\frac{p}{q}=\left|\frac{\log _{a}(1-x)}{\log _{a}(1+x)}\right|=\left|\log _{(1+x)}(1-x)\right|$
Given $0 < x < 1$. $\log _{(1+x)} \frac{1}{1-x}>1$
Therefore, $\frac{p}{q}>1, p>q$