Find the all possible real solutions of $x^y=y^x$ Find the all possible real solutions of $$x^y=y^x$$
$x,y$ both are real numbers.
My attempt:I observed the following solutions
$x=2,y=4$
$x=4,y=2$
$y=x$
Is there any other possible solutions?
 A: Applying the logarithm to both sides gives
$$y \log x = x \log y,$$ and rearranging gives
$$\frac{\log x}{x} = \frac{\log y}{y}$$, so we get a nontrivial solution (i.e., one for which $x \neq y$) for any value assumed twice by $f(x) := \frac{\log x}{x}$. More explicitly, if $x$ and $y$ are distinct values such that $f(x) = f(y)$, then by construction we have a nontrivial solution $x^y = y^x$.
The function $f$ is obviously continuous, and 


*

*has a single root at $x = 1$,

*is strictly increasing on $(0, e)$ and strictly decreasing on $(e, \infty)$,

*has a unique maximum of $\frac{1}{e}$ at $x = e$, and 

*has limit $\lim_{x \to \infty} f = 0$.


We can thus conclude that the values $f$ takes on twice are precisely those in $(0, \frac{1}{e})$. Given such a value, it is generally difficult to find its two preimages explicitly*, but nonetheless we can parameterize the solutions as
$$(x, y) = \left(\left(1 + \frac{1}{u}\right)^u , \left(1 + \frac{1}{u}\right)^{u + 1}\right).$$
See $x^y = y^x$ for integers $x$ and $y$ for much more.
*In closed form, anyway, we can write down solutions in terms of the Lambert-W function, which appears on this site much more than it does in "real life"...
