## Task A-3.1.

Determine all real numbers $x, y$ for which the equalities

$$
x^{\log y}+\sqrt{y^{\log x}}=110 \quad \text { and } \quad x y=1000
$$

hold.
## Solution.

Notice that

$$
x^{\log y}=\left(10^{\log x}\right)^{\log y}=10^{\log x \log y}=\left(10^{\log y}\right)^{\log x}=y^{\log x}
$$

The first equation can be rewritten as

$$
y^{\log x}+\sqrt{y^{\log x}}-110=0
$$

Using the substitution $t=\sqrt{y^{\log x}}$, we get the quadratic equation $t^{2}+t-110=0. \quad 1$ point The solutions to this quadratic equation are $t_{1}=-11, t_{2}=10$. Since $t>0$ must hold, the only solution is $t=10$.

From $\sqrt{y^{\log x}}=10$ it follows that $y^{\log x}=100$. Taking the logarithm, we have $\log x \log y=\log 100=2$. $\quad 2$ points Taking the logarithm of the second equation, we get $\log x+\log y=\log 1000=3$. 1 point

Thus, $\log x=3-\log y$, and substituting into the previous equation gives $(3-\log y) \log y=2$, or $-(\log y)^{2}+3 \log y-2=0$.

Using the substitution $u=\log y$, we get the quadratic equation $-u^{2}+3 u-2=0$. The solutions to this quadratic equation are $u_{1}=1, u_{2}=2$.

Substituting back, we have:

- $\log y=1: y=10, \log x=3-\log y=2, x=100$
- $\log y=2: y=100, \log x=3-\log y=1, x=10$.

Therefore, the solutions are $(x, y)=(10,100)$ and $(x, y)=(100,10)$.

1 point

Note: Finding both solutions is worth 1 point.