2. Solve the inequality $\log _{3}(x+2) \geq \log _{x+2} 3$.
Solution. Let $a=\log _{3}(x+2)$ and $b=\log _{(x+2)} 3$. Then $x+2=3^{a}$ and $3=(x+2)^{b}$, i.e., $x+2=(x+2)^{a b}$. According to this, $a b=1$, i.e., $b=\frac{1}{a}$. From $a \geq b$ it follows that $a \geq \frac{1}{a}$. For $a>0$, from $a^{2} \geq 1$ it follows that $a \geq 1$. Then $x+2=3^{a} \geq 3$, i.e., $x \geq 1$. For $a<0$, from $a^{2} \leq 1$ it follows that $-1 \leq a<0$. Then from $x+2=3^{a}$ we get that $3^{-1} \leq x+2<1$, i.e., $-\frac{5}{3} \leq x<-1$. Therefore, the solutions to the inequality are $x \geq 1$ or $-\frac{5}{3} \leq x<-1$.