# Solution $(x,y)$ to $x^y=y^x=a$, $a>1$ being real.

I seem to have found a method to compute the solution $(x,y)$ to the equation, $x^y=y^x=a$ $a\geq 1$ where $a$ is real, by using limits. But I don't know if this is something new. Does there already exist any method of finding such a solution? If yes, what is it?

• Maybe duplicate $x^y=y^x$
– GAVD
Commented Oct 2, 2015 at 3:00
• Take logarithm of both sides, and contemplate the result. Commented Oct 2, 2015 at 3:08
• @lubin ya we get x/y=logx/logy ..... how do we proceed after this in this method>? Commented Oct 7, 2015 at 18:52

From your equation $x_1/x_2=\log x_1/\log x_2$, you get $$\frac{\log x_1}{x_1}=\frac{\log x_2}{x_2}\,.$$ In other words, assuming that $x_1\ne x_2$, you’ll have two points of the same height on the graph $y=f(x)=(\log x)/x$. This function $f$ has derivative $f'(x)=(1-\log x)/x^2$, zero when $x=e$, nowhere else. So there’ll be a pair $(x_1,x_2)$ solving your equation with one of them greater than $e$, the other less. Since the maximum value of $f$ is $1/e$, all you need to do is choose any number $\alpha<1/e$ and find the two solutions of $f(x)=\alpha$. My primitive way of doing this would be numerical, using Newton-Raphson, I suppose. This method may be less elegant than some of the others given to the prior question, but it does show immediately that $(2,4)$ is the only solution with $x_1$ and $x_2$ integers and $x_1<x_2$.