# About $x^y=y^x$

Consider the equation $x^y=y^x$ and it is required to find ALL solutions $(x,y)$ for this equation,

It is obvious that all ordered pairs $(x,x)$, i.e., $y=x \ne 0$ is a trivial set of solutions. (Here, I am not sure if $x=y=0$ can be considered as a solution!!)

In addition, the set $x=\alpha^{\frac{1}{\alpha-1}},y=\alpha x$, for some real (not sure if complex works) value of $\alpha \ne 1,0$, would also fit in as solutions.

Are there any solutions which do not fit into the above two categories? Is it possible to prove otherwise?

Note by Janardhanan Sivaramakrishnan
5 years, 2 months ago

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There are solutions that do not fit. There will be a negative real solution as long as $y=\frac{a}{b}>1$, and there might be 2 negative real solutions when $y=\frac{a}{b}<1$, where $a$ is an even number and $b$ is an odd number. I haven't actually studied this in detail yet.

From what I got so far, numerically, there would be 2 negative real solutions when $y=\frac{a}{b}$, $k_c, $0.752688172043

- 5 years, 2 months ago

I do not even restricting $x,y$ to be real numbers.

Take $\alpha = -1$, we get $x=i=e^{\frac{\pi}{2}i},y=-i=e^{\frac{-\pi}{2}i}$ and

thus $x^y=e^{\frac{\pi}{2}i \times (-i)}=e^{\frac{\pi}{2}}$

and also $y^x=e^{\frac{-\pi}{2}i \times (i)}=e^{\frac{\pi}{2}}$

- 5 years, 2 months ago

Related: soumava's algorithm

- 4 years, 11 months ago

$\Large 2^4=4^2$

- 5 years, 2 months ago

This also fits the pattern. Take $\alpha = 2$, $x=2^{\frac{1}{2-1}}=2,y=2*2=4$.

- 5 years, 2 months ago