Solve the inequality $\log _{a}(x-a)>\log _{1 / a}(x+a)$.
Given $a>0$ and $a \neq 1$, and since $x-a>0$, it follows that $x>a$. Using the relationship between logarithms with bases $a$ and $\frac{1}{a}$, the inequality can be rewritten as:

$$
\log _{a}(x-a)>-\log _{a}(x+a)=\log _{a}(x+a)^{-1}=\log _{a} \frac{1}{x+a}
$$

If $a>1$, then the logarithmic function is strictly increasing, so in this case,

$$
x-a>\frac{1}{x+a}
$$

which implies $x^{2}-a^{2}>1$, and $x>\sqrt{a^{2}+1}(x>a)$ is the solution.

If $0<a<1$, then the logarithmic function is strictly decreasing, and

$$
x-a<\frac{1}{x+a}
$$

so $a<x<\sqrt{a^{2}+1}$ is the solution.