Task B-3.1. Solve the system of equations:

$$
\begin{aligned}
& 2^{x} \cdot 3^{y}=24 \\
& 2^{y} \cdot 3^{x}=54
\end{aligned}
$$
Solution. For the idea of multiplication or division or one of them.

By multiplying the given equations, we get

$$
\begin{aligned}
& 2^{x+y} \cdot 3^{x+y}=24 \cdot 54 \\
& 6^{x+y}=\left(2^{3} \cdot 3\right) \cdot\left(2 \cdot 3^{3}\right) \\
& 6^{x+y}=2^{4} \cdot 3^{4} \\
& 6^{x+y}=6^{4} \\
& x+y=4
\end{aligned}
$$

By dividing the given equations, we get

$$
\begin{aligned}
& 2^{x-y} \cdot 3^{x-y}=\frac{24}{54} \\
& \left(\frac{2}{3}\right)^{x-y}=\frac{2^{3} \cdot 3}{2 \cdot 3^{3}} \\
& \left(\frac{2}{3}\right)^{x-y}=\left(\frac{2}{3}\right)^{2} \\
& x-y=2
\end{aligned}
$$

(1 point)

From (1) and (2) it follows that

$$
x=3, \quad y=1
$$