How can we prove for the solutions of the equation $x^y=y^z=z^a...$? How can we prove for the solutions of the equation $x^y=y^z=z^a...$? You could natural log all of them and then proceed but I’m kind of confused. How would you solve for the solutions of a simpler case then, for example $x^y=y^z=z^x$? Since I think the problem above might be too hard...
 A: Consider your simpler case of
$$x^y = y^z = z^x \tag{1}\label{eq1}$$
If any of $x$, $y$ or $z$ is $1$, say $x = 1$, then $z^x = 1$ with $x = 1$ only if $z = 1$ and, likewise, $y^z = 1$ with $z = 1$ only if $y = 1$, so all $3$ values would be $1$. In general, $x = y = z$ is always a solution.
For the case where none of the values are $1$ and not all equal to each other, I assume that $x$, $y$ and $z$ are positive real numbers (I'll leave the cases where any of them may be negative to you) so you can then take the logarithm to any base $\gt 1$ (say the natural one, i.e., to the base $e$) of each value to get
$$y\ln x = z\ln y = x\ln z \tag{2}\label{eq2}$$
Since $\ln a \lt 0 \iff a \lt 1$ and $\ln a \gt 0 \iff a \gt 1$, this shows that all $3$ values of $x, y, z$ are either $\lt 1$ or $\gt 1$. First, assume they are all $\gt 1$ so the values in \eqref{eq2} are all positive. Due to symmetry, WLOG let
$$y \gt z \tag{3}\label{eq3}$$
Thus, \eqref{eq3} shows from the first & second parts of \eqref{eq2} that, since the logarithm function is strictly increasing, that
$$\ln y \gt \ln x \implies y \gt x \tag{4}\label{eq4}$$
Next, \eqref{eq4} means the first & third parts of \eqref{eq2} gives that
$$\ln z \gt \ln x \implies z \gt x \tag{5}\label{eq5}$$
This latest inequality of \eqref{eq5} means the second & third parts of \eqref{eq2} gives
$$\ln z \gt \ln y \implies z \gt y \tag{6}\label{eq6}$$
However, \eqref{eq6} contradicts \eqref{eq3}! Note you can use similar arguments for the case where all $x$, $y$ and $z$ are less than $1$, so the logarithm values are negative.
Also, you can extend this cyclic type argument to additional odd number of values & equations, which I'll leave up to you to do. However, note if the number of values is even, then there are exceptions, e.g., where the values alternate between $2$ and $4$, such as $2^4 = 4^2 = 2^4 = 4^2$.
