1. Determine the real solutions of the equation:

$$
\log _{2}(1+\sqrt{x})=\log _{3} x
$$
Solution. We introduce the substitution $\log _{3} x=y$, from which we get $\sqrt{x}=\sqrt{3^{y}}$. Substituting into the original equation, we obtain

$$
\begin{aligned}
& \log _{2}\left(1+\sqrt{3^{y}}\right)=y \\
& 1+\sqrt{3^{y}}=2^{y} \\
& \left(\frac{1}{2}\right)^{y}+\left(\frac{\sqrt{3}}{2}\right)^{y}=1
\end{aligned}
$$

An obvious solution to the last equation is $y=2$. If $y \geq 2$, then

$$
\left(\frac{1}{2}\right)^{y}\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \text { and }\left(\frac{\sqrt{3}}{2}\right)^{y}>\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{3}{4}
$$

from which

$$
\left(\frac{1}{2}\right)^{y}+\left(\frac{\sqrt{3}}{2}\right)^{y}>\frac{1}{4}+\frac{3}{4}=1
$$

Thus, $y=2$ is the only solution to the equation $\left(\frac{1}{2}\right)^{y}+\left(\frac{\sqrt{3}}{2}\right)^{y}=1$, from which, by returning to the substitution, we get that $x=9$ is the only solution to the original equation.