Solve the following equation and system of equations:

$$
x^{x^{\sqrt{x}}}=x^{\sqrt{x^{x}}}
$$

$$
\begin{array}{r}
y=m x \\
{ }^{b} \log y=m \cdot{ }^{b} \log x
\end{array}
$$
a) In the equation, $x$ cannot be negative, because then $\sqrt{x}$ would not be real, - and it cannot be 0 either, because then $0^{0}$ would appear in the exponents, whereas the zeroth power of 0 is only defined for bases different from 0. Therefore, $x>0$. - Furthermore, we see that $x=1$ is a solution to the equation (the value of both sides is 1), so in the search for other possible roots, we can assume that $x \neq 1$. - Due to the equality of the bases on both sides, the exponents can only be equal if the exponents are equal: $x^{\sqrt{x}}=\sqrt{x}^{x}$. Based on the identity $x=(\sqrt{x})^{2}$ valid for non-negative $x$, transforming the left side: $\sqrt{x}^{2 \sqrt{x}}=\sqrt{x}^{x}$. Since $\sqrt{x} \neq 1$, applying the above thought again, $2 \sqrt{x}=x$. From this, $4 x=x^{2}$ and the equation is satisfied only by $x=4$ among positive numbers. With this, the exponent on the left side is $4^{\sqrt{4}}=16$, and on the right side: $2^{4}=16$, so $x=4$ is also a root (the value of both sides is 4294967296).

b) The second equation of the system only makes sense if $x, y$ are positive, given that $b>0$ and $b \neq 1$; hence $y=x^{m}$. Eliminating $y$ gives $x^{m}=m x, x^{m-1}=m$, and finally

![](https://cdn.mathpix.com/cropped/2024_05_02_e210e0fe8918cccc96f7g-1.jpg?height=81&width=766&top_left_y=583&top_left_x=669)

The square root here makes no sense if $m-1=0$, i.e., $m=1$. We do not say there is no solution, because in this case both equations imply $y=x$, which is satisfied by any pair of equal positive numbers. There is no unique solution. - The equation (1) makes no sense for $m=0$ either, because $0^{-1}$ is not defined (0 only has positive exponents defined). Finally, from $x>0, y>0$ and the first equation $m=y / x$, there is no solution if $m<0$. - Therefore, (1) is a solution to the given system of equations if $b$ and $m$ are positive numbers different from 1.

Fritz József (Mosonmagyaróvár, Kossuth L. g. II. o. t.)

Remarks. 1. The root $x=1$ of the equation in a) can also be obtained through a "computational" approach from the equality of the logarithms of both sides: $x^{\sqrt{x}} \lg x=\sqrt{x}^{x} \lg x$ leads to $\left(x^{\sqrt{x}}-\sqrt{x}^{x}\right) \lg x=0$, which holds if $\lg x=0$, i.e., $x=1$.

2. It is useful to verify the solution for a few numerical values of $m$; let $b=10$ throughout. For example, with $m=2$: $x=2$, $y=4$; with $m=3$: $x=\sqrt{3}, y=3 \sqrt{3}$; with $m=10$: $x=\sqrt[9]{10}=1.292, y=12.92$; with $m=1/2$: $m-1=-1/2$, $1/(m-1)=-2$, $x=(1/2)^{-2}=2^{2}=4, y=2$, which is a swap of $x$ and $y$ compared to $m=2$, which is almost expected based on the first equation. The same is found for $m=1/3$ and $m=1/10$. Replacing $m$ with $1/\mu$ changes $1/(m-1)$ to $\mu/(1-\mu)$ and $m/(m-1)$ to $1/(1-\mu)$, so

$$
x=\left(\frac{1}{\mu}\right)^{\frac{\mu}{1-\mu}}=\mu^{\frac{\mu}{\mu-1}}, \quad y=\left(\frac{1}{\mu}\right)^{\frac{1}{1-\mu}}=\mu^{\frac{1}{\mu-1}}
$$

This shows the general validity of our observation. Numerical examples can lead to generally valid theorems if we dare to try non-integer numbers. Of course, it would be misleading to make general statements based only on a few numerical examples.