Solve the following equation:

$$
x^{x^{\frac{1}{2}}}=\frac{1}{2} .
$$
I. solution. The domain of the equation is the set of positive numbers. By substitution, we can verify that $\frac{1}{16}$ and $\frac{1}{4}$ are solutions to the equation. We will show that there are no other real solutions.

Taking the natural logarithm of both sides, we get an equivalent equation, since $x$ is positive:

$$
\sqrt{x} \ln x = -\ln 2
$$

Consider the function $f(x) = \sqrt{x} \ln x$ for $(x > 0)$. The function $f$ is differentiable on the set of positive numbers, and the derivative of the function is $f'(x) = \frac{\ln x + 2}{2 \sqrt{x}}$. If $0 < x < e^{-2}$, then $f'(x) < 0$; if $x > e^{-2}$, then $f'(x) > 0$. Therefore, $f$ is strictly decreasing if $0 < x < e^{-2}$ and strictly increasing if $x > e^{-2}$.

For this reason, $f$ can take a given value at most twice; there are no more solutions beyond the ones found; the problem has two solutions: $x_1 = \frac{1}{16}$ and $x_2 = \frac{1}{4}$.

Baharev Ali (Vác. Boronkay Gy. Gymn., 11th grade)

Note. In the correct solutions, various substitutions were used to simplify the function to be examined. Some examples: with the substitution $y = \sqrt{x}$, we get the equation $y^y = \frac{\sqrt{2}}{2}$; if $x = 2^k$, then the equation is $k \cdot 2^{\frac{k}{2}} = -1$. The following solution surprisingly leads to a simple equation.

II. solution. If $x > 0$, then there is exactly one $t \in \mathbb{R}$ such that $x = 2^{-2t}$. With this substitution, the equation takes the form

$$
\left(2^{-2t}\right)^{2^{-t}} = \frac{1}{2}
$$

Taking the base-2 logarithm of both sides:

$$
\begin{gathered}
2^{-t} \cdot \log_2 2^{-2t} = -1, \quad \text{i.e.,} \\
-t \cdot 2^{-t+1} = -1, \quad \text{or} \\
t = 2^{t-1}
\end{gathered}
$$

The function on the right side is convex, so a line can intersect its graph at most twice (figure). If $t = 1$ and $t = 2$, then we get solutions; hence $x_1 = 2^{-2} = \frac{1}{4}$, $x_2 = 2^{-4} = \frac{1}{16}$.

The solutions to the equation are therefore $\frac{1}{4}$ and $\frac{1}{16}$.

Hargita Gábor (Ócsa, Bolyai J. Gymn., 10th grade)

Note. Venter György (Fazekas M. Főv. Gymn., 12th grade) solves the problem with the substitution $x = \frac{1}{2^n A} \quad (n \geq 0$ integer, $1 \leq A < 2)$. First, he proves that $n \leq 5$, then finds the two solutions by case analysis.

![](https://cdn.mathpix.com/cropped/2024_05_02_d12f7a684d2c45b299c2g-1.jpg?height=510&width=521&top_left_y=1805&top_left_x=802)