0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $x = y$.

$\endgroup$
3
$\begingroup$

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 $[1,\infty)$.

$\endgroup$
1
$\begingroup$

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 !)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.