3. Solve the inequality $\log _{\frac{3 x-6}{3 x+5}}(3 x-9)^{10} \geq-10 \log _{\frac{3 x+5}{3 x-6}}(3 x+6)$.

Answer. $x \in\left(-2 ;-\frac{5}{3}\right) \cup(2 ; 3) \cup(3 ;+\infty)$.
The domain of definition is given by the inequalities $\frac{3 x-6}{3 x+5}>0, \quad \frac{3 x-6}{3 x+5} \neq 1, \quad 3 x+6>0, \quad 3 x-9 \neq 0,$ from which $\mathrm{x} \in\left(-2 ;-\frac{5}{3}\right) \cup(2 ; 3) \cup(3 ;+\infty)$. On the domain of definition, the inequality is equivalent to the following:

$$
\begin{aligned}
& \log _{\frac{3 x-6}{3 x+5}}(3 x-9)^{10} \geq \log _{\frac{3 x-6}{3 x+5}}(3 x+6)^{10} \Leftrightarrow\left(\frac{3 x-6}{3 x+5}-1\right)\left((3 x-9)^{10}-(3 x+6)^{10}\right) \geq 0 \Leftrightarrow \\
& \Leftrightarrow \frac{-11}{3 x+5}\left((3 x-9)^{2}-(3 x+6)^{2}\right) \geq 0 \Leftrightarrow \frac{-11}{3 x+5}(6 x-3)(-15) \geq 0 \Leftrightarrow x \in\left(-\infty ;-\frac{5}{3}\right) \cup\left[\frac{1}{2} ;+\infty\right) \text {. }
\end{aligned}
$$

Taking into account the domain of definition, we get $x \in\left(-2 ;-\frac{5}{3}\right) \cup(2 ; 3) \cup(3 ;+\infty)$.