Solve the following equation:

$$
x^{\sqrt{x}}=(\sqrt{x})^{x} .
$$
I. solution: We need to exclude the value $x=0$, because the expression $0^{0}$ is undefined. (Thus, 0 is not a solution to the equation even if the same undefined expression would stand on both sides after substitution.)

Raise both sides to the power of 2:

$$
x^{2 \sqrt{x}}=(\sqrt{x})^{2 x}=x^{x}
$$

Equal bases of powers can only be equal if their exponents are also equal, or if the base is 1. In the first case:

$$
2 \sqrt{x}=x, \quad \text { from which } \quad x=4
$$

In the second case:

$$
x=1
$$

By substitution, we can verify that both values are solutions to the equation.

Szatmári Gábor (Bp. VIII., Piarist g. II. y. t.)

II. solution: The value $x=0$ must be excluded. If we also consider that (for real solutions) the presence of $\sqrt{x}$ requires $x>0$, then we can take the logarithm of both sides:

$$
\sqrt{x} \cdot \lg x=x \lg \sqrt{x}=\frac{x}{2} \lg x
$$

that is,

$$
\lg x\left(\sqrt{x}-\frac{x}{2}\right)=\lg x(2-\sqrt{x}) \frac{\sqrt{x}}{2}=0
$$

This can only be true, since $x \neq 0$, if either the first factor is 0, so

$$
x=1
$$

or the second factor, then

$$
x=4
$$

Füle Károly (Bp. V., Apáczai Csere g. II. y. t.)