Does $x^x=y^y\iff x=y$, given $x,y\in\Bbb R$? I am curious whether the following is true:

Does $x^x=y^y\iff x=y$, given $x,y\in\Bbb R$?

I can see how $x=y\implies x^x=y^y$. We can raise one side of the equation by $x$ and another by $y$ and still maintain equality because $x=y$.
But the question becomes less clear when going the other way($x^x=y^y\implies x=y$). It is not intuitive whether or not this is true, and I can't seem to see any counterexample to this. I tried to take the $\ln$ of both sides to get $x\ln x=y\ln y$, and maybe do $e$ to the power of both sides(to get rid of $\ln$), but I'm not too sure if I'm on the right track. Also, $\ln$ is only valid for $x,y>0$ so it doesn't even cover all the possibilities. This question might be trivial, but I don't really see how to continue from here. Any thoughts?
 A: Consider the graph of $f(x)=x^x$:

This graph appears to have a turning point somewhere around $x=0.5$, and we can use calculus to find out exactly when this is true. If $z=x^x$, then $\ln z = \ln x^x=x\ln x$. If we differentiate both sides of this equation, we get
\begin{align}
\frac{1}{z}\frac{dz}{dx}&=\ln x + x\cdot\frac{1}{x}=\ln x + 1 \\[6pt]
\frac{dz}{dx}&=z(\ln x + 1)
\end{align}
To find the turning point set $\frac{dz}{dx}$ equal to $0$. It turns out that the graph has a turning point when $x=1/e$. Of course, in this case, what matters is not when the graph has a turning point, but the fact that it does have one. This means that $f$ is not one-to-one, i.e. there may be multiple $x$ that correspond to the same output. Can you see where this is going?
A: In addition to there being a range of real solutions for $x^x=y^y, x\not=y$, there are also infinitely many solutions where both $x$ and $y$ are rational.
For any positive whole number $n$, render
$x=(n/(n+1))^n, y=(n/(n+1))^{n+1}<x; x,y\in\mathbb{Q}$
Then
$y^y=\left(\left(\dfrac{n}{n+1}\right)^{n+1}\right)^{\left(\dfrac{n}{n+1}\right)^{n+1}}$
$=\left(\left(\dfrac{n}{n+1}\right)^{n+1}\right)^{\left(\left(\dfrac{n}{n+1}\right)\left(\dfrac{n}{n+1}\right)^n\right)}$
$=\left(\left(\dfrac{n}{n+1}\right)^{(n+1)\left(\dfrac{n}{n+1}\right)}\right)^{\left(\dfrac{n}{n+1}\right)^n}$
$=\left(\left(\dfrac{n}{n+1}\right)^n\right)^{\left(\dfrac{n}{n+1}\right)^n}=x^x$
With $y<x,x,y\in\mathbb{Q}$ we have an all-rational solution for each positive whole number $n$.  Combined with the abalysidls by Joe this result implies more precisely $y<e<x$.
A: We have $x \log(x) = y \log(y)$. Because we are seeking for solutions with $x\ne y$ and the equation is symmetrix, assume WLOG that $y>x$. Then let $y= a x$ for some $a>1$. Then
$$\begin{align}
 x \log(x) &= ax \log(x) + ax \log(a)\\
 x\log (x) (1-a) &= ax \log(a) \\
x  &=  a^{a/(1-a)} 
\end{align}
$$
For $a\to 1$ this tends to $x_0 = 1/e$.
Then we have infinite pair of solutions with $x \in (0, 1/e)$  and $y \in (1/e, 1)$
Some values
a        x        y
1.01   0.36605   0.36971
2      1/4       1/2
5      0.13375   0.66874 

