3. Solve the inequality

$$
\log _{3} x \geq \log _{x} 81
$$
Solution. The domain of definition of $\log _{3} x$ and $\log _{x} 81$ is $(0,1) \cup(1,+\infty)$, so the set of solutions to the given inequality must be a subset of that domain. The given inequality is transformed into:

$$
\log _{3} x \geq \frac{4}{\log _{3} x}
$$

There are two cases.

(i) $\log _{3} x>0$, i.e., $x>1$. Then the inequality is transformed into $\log _{3} x \geq 2$, and from here we get that $x \geq 3^{2}=9$. The set of solutions in this case is $[9,+\infty)$.

(ii) $\log _{3} x<0$, i.e., $0<x<1$. Then the inequality is transformed into $\log _{3} x \geq-2$, and from here we get that $x \geq \frac{1}{9}$. The set of solutions in this case is $\left[\frac{1}{9}, 1\right)$.

Therefore, the set of solutions to the given inequality is $\left[\frac{1}{9}, 1\right) \cup[9,+\infty)$.