3. Solve the system of equations

$$
\begin{cases}x^{y} & =y^{x} \\ a^{x} & =b^{y}\end{cases}
$$

$a \neq 1, b \neq 1$. Perform a discussion.
Solution. The solutions of the system can only be $x>0$ and $y>0$. From this and the second equation, it follows that $a>1$ and $b>1$, or $0<a<1$ and $0<b<1$. If $a>1$ and $0<b<1$, or $0<a<1$ and $b>1$, then the system has no solution.

Therefore, let $a>1$ and $b>1$ (or $0<a<1$ and $0<b<1$), and $x>0$, $y>0$, we obtain the system

$$
\begin{gathered}
\log a \cdot \log x=\log b \cdot \log y \\
x \log a=y \log b
\end{gathered}
$$

which is also equivalent to the given system. From the first equation of the last system, we find that

$$
x=y^{\frac{\log b}{\log a}}
$$

substituting into the second equation:

$$
y^{\frac{\log b-\log a}{\log a}}=\frac{\log b}{\log a}
$$

From this, if $\log b-\log a \neq 0$, we have

$$
y=\left(\frac{\log b}{\log a}\right)^{\frac{\log a}{\log b-\log a}}, x=\left(\frac{\log b}{\log a}\right)^{\frac{\log b}{\log b-\log a}}
$$

If, however, $\log b-\log a=0$, then $a=b$ and from the second equality we get $x=y$. In that case, the system is satisfied by any pair of numbers $x$ and $y$ equal to each other.