I found a graph of the solutions to the equation $x^y = y^x$ online; the one who posted the graph was curious as to why the set of solutions formed a line and a curve that intersect at the point $(e,e)$. The line of solutions is easy to understand - it is the line $y = x$. Finding an equation for the curve is a bit more aggravating.
To start, we manipulate the equation $x^y = y^x$ until we obtain $$\frac{y}{x} = \frac{\ln{y}}{\ln{x}}$$ To make things a little simpler, we concern ourselves with the first quadrant of $\mathbb{R}^2$ and define $C = \frac{y}{x}$, noting that $C > 0$. This yields the pair of equations $$ y = Cx$$ $$ y = x^C$$ With some simple manipulations, we find $$x = C^{\frac{1}{C-1}}$$ $$y = C^{\frac{C}{C-1}}$$ There is a slight reason for worry when $C = 1$, but we recall $$\lim_{C\to1} C^{\frac{1}{C-1}} = \lim_{C\to1} C^{\frac{C}{C-1}} = e$$ which is exactly what we want. In summary, for any given $C \in \mathbb{R}^+\setminus\{1\}$ we can find a point on the curve (under our restriction).
Unfortunately, I was not able to go much further. Is there a function that models the curve of solutions to $x^y = y^x$, i.e. the set $$S = \left\{\left(C^{\frac{1}{C-1}},C^{\frac{C}{C-1}}\right) : C \in \mathbb{R}^+\setminus\{1\}\right\} \cup \{(e,e)\}$$ If so, how do we derive it?
