1A. In the set of real numbers, solve the equation

$$
x^{\log _{3}(x-1)}+2(x-1)^{\log _{3} x}=3 x^{2}
$$
Solution. Clearly, $x \in(1,+\infty)$. Let $\log _{3}(x-1)=u$ and $\log _{3} x=v$. We have

$$
\begin{aligned}
& x^{\log _{3}(x-1)}=3^{\log _{3} x^{\log _{3}(x-1)}}=3^{\log _{3}(x-1) \cdot \log _{3} x}=3^{u v} \text { and } \\
& (x-1)^{\log _{3} x}=3^{\log _{3}(x-1)^{\log _{3} x}}=3^{\log _{3}(x-1) \cdot \log _{3} x}=3^{u v}
\end{aligned}
$$

Furthermore, $x^{2}=3^{\log _{3} x^{2}}=3^{2 \log _{3} x}=3^{2 v}$, so the given equation becomes $3 \cdot 3^{u v}=3 \cdot 3^{2 v}$, from which it follows that $u v=2 v$, i.e., $v=0$ or $u=2$. If $v=0$, then from $\log _{3} x=v$ it follows that $x=1$, which is not possible since $x \in(1,+\infty)$. If $u=2$, then from $\log _{3}(x-1)=u$ it follows that $x=10$ and this is the only solution to the given equation.