Solve the following system of equations:

$$
\begin{aligned}
x^{(\lg y)^{\lg \lg x}} & =10^{y^{2}} \\
y^{(\lg x)^{\lg \lg y}} & =y^{y}
\end{aligned}
$$
I. Solution: Since the lg function is not defined for 0 and negative values, it must be that $x>0$ and $y>0$. Furthermore, if $0<1$ and (for similar reasons) $y>1$.

The equation (1) is transformed by taking the logarithm twice. (This is permissible because, according to our conditions, the quantities to be logarithmed are all positive.)

$$
\begin{aligned}
& (\lg y)^{\lg \lg x} \cdot \lg x=y^{2} \\
& (\lg \lg x)(\lg \lg y)+\lg \lg x=2 \lg y
\end{aligned}
$$

According to our condition, $y \neq 0$ and $y \neq 1$, so in equation (2), the exponent of $y$ on both the left and right sides is equal.

$$
(\lg x)^{\lg \lg y}=y
$$

Let's take the lg of both sides of this last equation.

$$
(\lg \lg y)(\lg \lg x)=\lg y
$$

Subtracting this from ( $1^{\prime \prime}$ )

$$
\lg \lg x=\lg y
$$

Substituting this back into $\left(2^{\prime \prime}\right)$ and dividing by $\lg y$ (since $\lg y \neq 0$)

$$
\lg \lg y=1, \quad \text { from which } \quad \lg y=10, \quad \text { and thus } \quad y=10^{10} \text {. }
$$

However, from (3) $\lg x=y$, so

$$
x=10^{y}=10^{10^{10}}
$$

This is the only solution to our system of equations.

József Szabados (Bp. III. Árpád g. III. o. t.)

II. Solution: The exponents on the left sides are equal, which we can see clearly if we compare the lg of the exponents. Knowing this, let's take the lg of both equations, and divide the two resulting equations by each other:

$$
\begin{gathered}
(\lg y)^{\lg \lg x} \cdot \lg x=y^{2} \\
(\lg x)^{\lg \lg y} \cdot \lg y=y \lg y \\
\frac{\lg x}{\lg y}=\frac{y^{2}}{y \lg y}, \quad \text { or } \quad \lg x=y
\end{gathered}
$$

and with this we arrive at equation (3) of the first solution, and substituting $y$ for $\lg x$ in $\left(2^{*}\right)$, $y$ can be calculated in the same way as in the first solution.

László Farkas (Ózd, József A. g. III. o. t.)