Solve the following system of equations:

$$
\begin{aligned}
x^{y} & =y^{x} \\
100^{x} & =500^{y}
\end{aligned}
$$
$$
\begin{aligned}
x^{y} & =y^{x} \\
100^{x} & =500^{y}
\end{aligned}
$$

If we take the logarithm of both sides of the equations, we obtain the following system of equations:

$$
\begin{gathered}
y \log x=x \log y \\
y \log 100=y \log 500
\end{gathered}
$$

By multiplying the left and right sides of these equations, the following equation holds:

$$
2 \log x=(2+\log 5) \log y
$$

Taking the logarithm of the terms in equation 4), we obtain the following equation:

$$
\log x=\log y+\log \frac{1}{2}(2+\log 5)
$$

If we now eliminate $\log x$ from equations 5) and 6),

$$
\log y=\frac{2 \log z}{\log 5}
$$

| $\log 5=0.69897$ | $0.260340: \underline{0.69897}$ |
| :---: | :---: |
| $\frac{1}{2} \log 5=0.34949$ | 506490.37246 |
| $z=1.34949$ | 1721 |
| $\log z=0.13001$ | 323 |
| 13.2 | 44 |
| 2.97 | 3 |
| $\overline{\log z=0.13017}$ |  |
| $2 \log z=0.26034$ | $\log y=0.37246$ |
| 0.37246 | $\underline{36}$ |
| 0.13017 | 10 |
| $\overline{\log x=0.50263}$ | $\overline{y=2.3575}$ |
| $\underline{256}$ | $x=3.1815$ |

where

$$
z=1+\frac{1}{2} \log 5
$$

and thus from 6),

$$
\log x=\log y+\log z
$$

The further course of the calculation is now as follows:

7

(Aladár Visnya, real school, 8th grade, Pécs.)

The problem was also solved by: Béla Bolemann, evangelical gymnasium, 8th grade, Budapest; Gusztáv Grossmann, evangelical gymnasium, 8th grade, Budapest; Tivadar Heymann and József Rothschild, real school, 8th grade, Györ.