Prove that there exists infinitely many positive numbers $x$ and $y$ such that $x\neq y$ but $ x^x=y^y$ Prove that there exists infinitely many pairs of positive real numbers $x$ and $y$ such that $x\neq y$ but $ x^x=y^y$.
For example $\tfrac{1}{4} \neq \tfrac{1}{2}$ but 
$$\left( \frac{1}{4} \right)^{1/4} = \left( \frac{1}{2} \right)^{1/2}$$
I am confused how to approach the problem. I think we have to find all the sloutions in a certain interval, probably $(0,1]$.
 A: It is easier than that, you need only examine the derivative: the derivative is $x^x \left ( \frac{d}{dx} x \log(x) \right ) = x^x \left ( \log(x) + 1 \right )$ which changes sign precisely at $x=e^{-1}$. This point is a minimum, and of course $\lim_{x \to \infty} x^x=\infty$. Knowing that, do you see now how for every $x \in (0,e^{-1})$ there exists $y>e^{-1}$ with $x^x=y^y$?
(More generally the original statement holds whenever any continuous function has a local extremum.)
A: While I like the continuity-based approach of the other answers, you can also get a parameterized set of solutions through algebraic methods.
Let $x$ be the larger of the two and define $a \in (0,1)$ by $a = y/x$. Then
$$
x^x = y^y = (ax)^{ax}\,\,\, \Longrightarrow\,\,\, x = a^a x^a\Longrightarrow \,\,\,x = a^{a/(1-a)},
$$
and correspondingly $y = a^{1/(1-a)}$. Since there are infinitely many $a\in (0,1)$, there are infinitely many solutions to $x^x = y^y$
A: In fact is is true for any $a<b$ (in this case, $a=1/4,  b=1/2$) and any function $f$ satisfying


*

*$f(a)=f(b)$ and

*$f$ is continuous on $[a,b]$
that there are infinitely many pairs $x<y$ satisfying $f(x)=f(y)$. 
If $f$ is constant on $[a,b]$ this is trivially true; if not there is some $c\in [a,b]$ with $f(c)\neq f(a)$, and then the result follows by applying the intermediate value theorem to $[a,c]$ and $[c,b]$.
