Graph of the function $x^y = y^x$, and $e$ (Euler's number). Earlier, I was using the Desmos Graphing Calculator, and I wanted to remind myself of what the graph of the function $x^y = y^x$ looked like.
If you have never seen what it looks like before, it is similar to the shape of the Greek letter, psi (ψ); composed of two graphs:


*

*$y = x$

*I am unsure as to what the equation of this graph is equal to, but I would say that it is similar in shape to the graph of $y = \frac{1}{x}$, but much steeper.


However, what I do know is that these two graphs intercept each other at the point $(e, e)$, where $e$ is Euler's number.
Please, if you could, check it out for yourself, but my question is, why is Euler's number related to this graph?
Edit:
I believe, upon further inspection, that it is related to Euler's number due to the fact that $x^y = e^{x \ln y}$, and, therefore, $y^x = e^{y \ln x}$.
Due to this fact, we can take the natural logarithm of both sides of the equation $e^{x \ln y} = e^{y \ln x}$, in order to get that $x \ln y = y \ln x$.
I do not know how to continue with this explanation, but my new question is, what is the equation of the graph that I could not work out?
Thank you, and good luck!
 A: For a given $y>0$, the equation $x\ln(y) = y\ln(x)$ is equivalent to $\ln(x)/x=\ln(y)/y$.
Here is the graph of the function $f(x)=\displaystyle\frac{\ln(x)}{x}$.

The function has a global maximum at $x=e$.
If $y\leq 1$, then $f(y)\leq0$ and the equation has only one solution: $x=y$.
If $y>1$, then $f(y)>0$ and $y\neq e$ and the equation has two solutions (you can see that an horizontal line will cut the graph of the function at two points, unless $y=e$, in which case the "two" solutions become equal.)
If $y=e$, then $x=e$ (This is the point where your two graphs intersect.)
I don't think there is a closed form for your second graph (that is, using only elementary functions.) But you can use the Lambert W function to get something close enough.
The Lambert W functions $(W_k)_{k \in \mathbb Z}$ are the solutions of the equation: $$\forall k \in \mathbb Z:\forall x\in \mathbb R: W_k(x)e^{W_k(x)}=x$$
If we want real valued solutions (which we do, here), then we need only $W_0$ and $W_{-1}$.
We have:
$$\frac{1}{x}\ln(x)=\frac{1}{y}\ln(y)$$
$$\frac{1}{x}\ln(\frac{1}{x})=\frac{1}{y}\ln(\frac{1}{y})$$
$$\ln(\frac{1}{x})e^{\ln(\frac{1}{x})}=\ln(\frac{1}{y})e^{\ln(\frac{1}{y})}$$
Therefore:
$$\ln(\frac{1}{y})=W_0(\frac{1}{x}\ln(\frac{1}{x}))\text{ or } \ln(\frac{1}{y})=W_{-1}(\frac{1}{x}\ln(\frac{1}{x}))$$
$$y=e^{-W_0(\frac{1}{x}\ln(\frac{1}{x}))}\text{ or } y=e^{-W_{-1}(\frac{1}{x}\ln(\frac{1}{x}))}$$
We simplify it a little to get the final result:
$$y=-\frac{xW_{0}(\frac{-\ln(x)}{x})}{\ln(x)}\text{ or } y=-\frac{xW_{-1}(\frac{-\ln(x)}{x})}{\ln(x)}$$
This gives us two graphs.


A superposition of the two graphs gives us the one you saw initially.

A: Any equation which can be written $$A+Bx+C\log(D+Ex)=0$$ has solutions expressed in terms of Lambert function.
In the case of $x^y=y^x$, this writes $$y = -\frac{x}{\log x}\,W\left(-\frac{\log x}{x}\right)$$ and what Lambert and Euler showed is that, in the real domain, the function $W(z)$ exists if $z\geq -\frac 1e$. So, for the argument $-\frac{\log x}{x}\geq -\frac 1e$, this implies $x \geq e$. This corresponds to the second branch of the curve (after $y=x$). At $x=e$, the slope of the curve is $-1$.
A: For most points on $y=x$, you have $dy/dx=1$.  There is one point where $dy/dx$ could be $1$ or $-1$.  Call that point $(x,y)=(a,a)$.
$$x\ln y=y\ln x\\
\ln y+\frac xy\frac{dy}{dx}=\frac{dy}{dx}\ln x+\frac yx\\
\ln a+\frac{dy}{dx}=\frac{dy}{dx}\ln a+1$$
This has a unique solution unless $\ln a=1$
A: Here is a relevant paper. They find a parametric equation for the non-$“x=y”$ branch of the graph (Desmos link), and then prove that the branches intersect at $(e,e)$. The rest of this answer is my own input.
The fact that the two branches of the graph of $x^y=y^x$ intersect at $(e,e)$ is related to the fact that the graphs of $y=x^e$ and $y=e^x$ are tangent. (See, we have two solutions for things like $x^3=3^x$, which is why there are two points with $y=3$ on that graph. However, there is only one solution to $x^e=e^x$, since they're tangent.)
Theorem: $x^e$ and $e^x$ are tangent.
Proof: Remember that
$$e^t\ge t+1\quad\text{for all }t$$
That's one of my favorite formulae. It lets us prove so many things about $e$ without calculus.
Substituting in $t=\dfrac xe-1$ gives us:
$$e^{x/e-1}\ge\dfrac xe$$
Multiplying by $e$:
$$e^{x/e}\ge x$$
Raising to the power of $e$:
$$e^x\ge x^e$$
Since they clearly intersect at $x=e$, they must be tangent.$\tag*{$\blacksquare$}$
P.S. That last equation solves the puzzle of which is larger, $e^\pi$ or $\pi^e$.
P.P.S. Extra credit: Using the fact that $e^t\ge t+1$, and without calculus, prove that $1-\frac12+\frac13-\dotsb=\ln2$. Warning: This is hard. (I say "no calculus," but you kind of need the Squeeze Theorem at the very end.)
A: This is a derivation of Claude's answer.
The definition of the Lambert-$W$ function is the solution to:
$$x = y e^y$$
Our equation is:
$$x \ln y = y \ln x$$
$$ \implies - \frac {\ln x}{x} = -\frac {\ln y}y = - \ln y e^{- \ln y}$$
$$\implies W\left({-\frac {\ln x}{x}}\right) = - \ln y = - y \frac{\ln x}{x}$$
$$\implies y = - \frac{x}{\ln x}W\left({-\frac {\ln x}{x}}\right) $$
A: Note that $y^x=x^y$ is symmetrical across $y=x$.  From here we can conclude that the curved part of your graph has a slope of $-1$ at the point in interest.  If it didn't, then it wouldn't be symmetrical.
We can prove it symmetrical across $y=x$ by taking the inverse function and noting the inverse function is indeed still equal to the original equation.
Upon solving for $y$ we get $$y=e^{-W(-\frac{\ln(x)}{x})}$$
Differentiate:$$y'=[\frac{\ln(x)-1}{x^2}][\frac1{-\frac{\ln(x)}{x}+e^{W(-\frac{\ln(x)}{x})}}]e^{-W(-\frac{\ln(x)}{x})}$$
Where setting it equal to $-1$, we get $x=e$.
I have no idea how you would solve for that, but that is a solution.
