2B. In the set of real numbers, solve the equation

$$
3^{\log _{x} 3} \cdot x^{\log _{3} x}=9
$$
Solution. The domain of the equation is $x>0$ and $x \neq 1$. We take the logarithm with base 3 and obtain the equivalent equations

$$
\begin{aligned}
& \log _{3}\left(3^{\log _{x} 3} \cdot x^{\log _{3} x}\right)=\log _{3} 9 \Leftrightarrow \\
& \log _{x} 3+\left(\log _{3} x\right)^{2}=2 \Leftrightarrow \\
& \frac{1}{\log _{3} x}+\left(\log _{3} x\right)^{2}=2
\end{aligned}
$$

We introduce the substitution $\log _{3} x=u$, i.e., $x=3^{u}$, so the equation takes the form $\frac{1}{u}+u^{2}-2=0 \Leftrightarrow u^{3}-2 u+1=0 \Leftrightarrow(u-1)\left(u^{2}+u-1\right)=0$. The solutions to this equation are $u_{1}=1, u_{2,3}=\frac{-1 \pm \sqrt{5}}{2}$, so the solutions to the original equation are $x_{1}=3$ and $x_{2,3}=3^{\frac{-1+\sqrt{5}}{2}}$. Moreover, $x_{1}, x_{2}, x_{3}$ belong to the domain.