Solve the following system of equations:

$$
\begin{aligned}
& x y=500 \\
& x^{\lg y}=25
\end{aligned}
$$
Since $\lg y$ appears in the second equation, $y$ can only be positive, but then $x$ must also be positive according to the first equation.

Therefore, it is free to take the logarithm of both sides of both equations:

$\lg x + \lg y = \lg 500 = 2 + \lg 5$

$\lg x \cdot \lg y = \lg 25 = 2 \lg 5$

Based on the relationship between the roots and coefficients of a quadratic equation, $\lg x$ and $\lg y$ are the two roots of the quadratic equation

$$
t^{2} - (2 + \lg 5) t + 2 \lg 5 = 0
$$

The values of these roots are

$$
\begin{aligned}
& \lg x_{1} = \lg y_{2} = 2 \\
& \lg x_{2} = \lg y_{1} = \lg 5
\end{aligned}
$$

From this,

$$
x_{1} = y_{2} = 100
$$

and

$$
x_{2} = y_{1} = 5
$$

Imre Csiszár (Bp. I., Petőfi g. III. о. t.)