If $x^x=y^y$ and $x,y>1$, prove that $x=y$

A friend and I have been trying to prove the following:

If $x^x=y^y$, where $x,y\ge1$, prove that $x=y$.

Intuitively, it's correct, but we haven't been able to prove it. We've tried a few different ways, most of them involving logarithms, but nothing has worked as of yet. I don't think logarithms are the correct way to do this. Any help would be greatly appreciated.

Note that we can assume that either $x > 1$ or $y > 1$ since the statement that we want to prove is true if both of them are equal to $1$. In the following, without loss of generality, we assume that $x > 1$ (we can always switch the role of $x$ and $y$): \begin{align} &&x^x &= y^y \\ \tag{Apply the $\log$ function on both sides.} \\ &\Rightarrow& x\log{x} &= y\log{y} \\ \tag{We have $y \geq 1$ and $x > 1 \Rightarrow \log{x} > 0$.} \\ &\Rightarrow& \frac{x}{y} &= \frac{\log{y}}{\log{x}} \\ \end{align} Now by contradiction assume that $x \neq y$ we have two cases:

• $x > y \Rightarrow \log{x} > \log{y} \Rightarrow \left\lbrace \begin{matrix}\frac{x}{y} > 1 \\ \frac{\log{y}}{\log{x}} < 1\end{matrix} \right.$ which is not possible since $\frac{x}{y} = \frac{\log{y}}{\log{x}}$.
• $x < y \Rightarrow \log{x} < \log{y} \Rightarrow \left\lbrace \begin{matrix}\frac{x}{y} < 1 \\ \frac{\log{y}}{\log{x}} > 1\end{matrix} \right.$ which is not possible since $\frac{x}{y} = \frac{\log{y}}{\log{x}}$.

From the two contradictions above we can deduce that $x \neq y$ is not possible and hence $x = y$.

The derivative of the function $$f(t)=t^t$$ is $$f'(t)=t^t(1+\ln t)$$ which is positive for $t\ge 1$. Then $f$ is increasing and hence injective in $[1,\infty)$.

Assume that $x>y$, then $x=y+h$ for some $h>0$. Then we get:

$x \ln x =(y+h) \ln(y+h) > y \ln y$ and therefore

$x^x=e^{x \ln x} > e^{y \ln y}=y^y$, a contradiction.

(All functions involved are strictly increasing !)