## Task A-4.1. (8 points)

Solve the equation $\sqrt{x^{x}}=x^{\sqrt{x}}$ in the set of real numbers.
## Solution.

First method.

From the equation, we see that $x>0$ must hold.

The given equation is equivalent to $x^{\frac{x}{2}}=x^{\sqrt{x}}$.

This equality holds if $x=1$ or if the exponents are equal.

From the equation $\frac{x}{2}=\sqrt{x}$, we get $x^{2}-4 x=0$, or $x=0, x=4$.

Due to the condition, the solutions to the original equation are only $x=1$ and $x=4$.

Second method.

From the equation, we see that $x>0$ must hold.

Since both sides are positive, we can take the logarithm: $\log x^{\frac{x}{2}}=\log x^{\sqrt{x}}$, which leads to $\frac{1}{2} x \log x=\sqrt{x} \log x$. From $\log x\left(\frac{1}{2} x-\sqrt{x}\right)$, we get two simple equations: $\log x=0$ and $\frac{1}{2} x=\sqrt{x}$, whose solutions are $x=1$ or $x=0$ and $x=4$.

Due to the condition, the solutions to the original equation are only $x=1$ and $x=4$.

For 7 points: omitting the solution $x=1$ or retaining the solution $x=0$.

For 8 points: correct result and procedure.