Problem 10.1. Solve the system

$$
\left\lvert\, \begin{aligned}
& 3 \cdot 4^{x}+2^{x+1} \cdot 3^{y}-9^{y}=0 \\
& 2 \cdot 4^{x}-5 \cdot 2^{x} \cdot 3^{y}+9^{y}=-8
\end{aligned}\right.
$$

Ivan Landjev
10.1. Set $u=2^{x}>0$ and $v=3^{y}>0$. Then the system becomes

$$
\left\lvert\, \begin{aligned}
& 3 u^{2}+2 u v-v^{2}=0 \\
& 2 u^{2}-5 u v+v^{2}=-8
\end{aligned}\right.
$$

The first equation can be written as $(u+v)(3 u-v)=0$, whence $u=-v$ or $3 u=v$. The first case is not possible since $u$ and $v$ are positive. Plugging $v=3 u$ in the second equation gives $u^{2}=2$, whence $u=\sqrt{2}$ and $v=3 \sqrt{2}$. Thus $x=\frac{1}{2}$ and $y=1+\frac{1}{2} \log _{3} 2$.