4A. Solve the system

$$
\left\{\begin{array}{l}
x^{\log y}+\sqrt{y^{\log x}}=110 \\
x y=1000
\end{array}\right.
$$
Solution. Note that $x > 0$ must hold. If we introduce the notations $A = x^{\log y}$, $B = y^{\log x}$, then $\log A = \log x \cdot \log y = \log B$, from which it follows that $A = B$. Therefore, we can introduce the substitution $x^{\log y} = y^{\log x} = t$. The first equation of the system then takes the form $t + \sqrt{t} = 110$, and with the substitution $\sqrt{t} = z$, where $t, z \geq 0$, we get the standard quadratic equation $z^2 + z - 110 = 0$ with solutions $z_1 = -11$ and $z_2 = 10$. Due to positivity, only $z = 10$ remains, i.e., $t = 100$. The original system can now be written in the form

$$
\left\{\begin{array}{l}
x^{\log y} = 100 \\
x y = 1000
\end{array}\right.
$$

We logarithmize both equations and get

$$
\left\{\begin{array}{l}
\log x \cdot \log y = 2 \\
\log x + \log y = 3
\end{array}\right.
$$

We can observe that $\log x$ and $\log y$, according to Vieta's formulas, are the roots of the quadratic equation $p^2 - 3p + 2 = 0$ whose solutions are $p_1 = 1, p_2 = 2$. Finally, from

$$
\left\{\begin{array}{l}
\log x = 1 \\
\log y = 2
\end{array} \text { or } \left\{\begin{array}{l}
\log x = 2 \\
\log y = 1
\end{array}\right.\right.
$$

we obtain the solutions of the system $(x, y) \in \{(10, 100), (100, 10)\}$.