Self intersection of the implicit curve $x^y-y^x=0$

Question

Watching the graph of the curve defined by $x^y=y^x$, which contains the line $y=x$, I noticed that the line intersects the curve itself only at one point that looks to be $(e;\;e)$.

How can I formally prove this?

Answer 1

HINT: fix $y>0$ and study the function $x\mapsto x^y-y^x$ and exploit the symmetry.

Answer 2

One approach is to notice that $x^y = y^x \implies \frac{\log x}{x} = \frac{\log y}{y}$ and look at the partial derivatives of $f(x,y)=\frac{\log x}{x} - \frac{\log y}{y}$. The partials at the point of intersection should be $0$ or undefined.

$$ \nabla f = \begin{bmatrix} \frac{1-\log x}{x^2} \\ \frac{\log y -1}{y^2} \end{bmatrix} ,$$

which is undefined for $x \le 0$ and $\nabla f = \textbf{0}$ when $x=e$.

A second approach would be to rewrite the original expression as $y=\frac{-x W(-\frac{\log x}{x})}{\log x}$, where $W(z)$ is the Lambert-$W$, and find where $\frac{dy}{dx} = 0$ or is undefined, which also happens to be $x \le 0$ and $x=e$.

Answer 3

HINT.-Your graph is a union of two curves, $x-y=0$ and another of an"hyperbolic" shape.What you have to do is plot (using the same function plotter you have used) the function $F(x,y)=\frac{x^y-y^x}{x-y}=0$ which is none other than the "hyperbolic" part of your curve. Taking the appropriate scale the plotter give you the corresponding intersection you are looking for which is in fact $(e,e)$ as you guess.