Solve $x^y \, = \, y^x$

The function $u:x\mapsto(\log x)/x$ is increasing on $[1,\mathrm e]$ from $u(1)=0$ to $u(\mathrm e)=1/\mathrm e$, and decreasing on $[\mathrm e,+\infty)$ from $u(\mathrm e)=1/\mathrm e$ to $0$. Hence, if $u(x)=u(y)$ with $y\gt x\geqslant 1$, then $y\gt \mathrm e\gt x\gt1$. Since $\mathrm e\lt3$, this implies that $1\lt x\lt3$. If furthermore $x$ is an integer, then $x=2$. The unique root of the equation $u(y)=(\log2)/2$ such that $y\gt\mathrm e$ is $y=4$. Hence $(x,y)=(2,4)$ is the unique solution.