Number of solutions for $x ^ y = y ^ x = k$

Question

For a given value of $k \geq 0$, how many solutions $x, y \in \mathbb R$ are there to $x ^ y = y ^ x = k$?

My attempt so far:

There is the "trivial" solution where $x = y$, and the problem reduces to $x ^ x = k$. This exists for any $k \geqslant 1$.

For $k < 1$, the situation gets weird. Between around $\ln 2$ and $1$, there are two solutions to $x ^ x = k$, and less than around $\ln 2$ there are no real solutions to it.

Based on my experimentation, there seems to be another inflection point at $k > e ^ e$ where suddenly a second solution shows up with $x \ne y$. For example, 16 has the solution (2, 4).

Specific sub-questions

• Why the inflection point at $\ln 2$?
• Why the inflection point around $ e ^ e $?
• Are there other inflection points?

Answer 1

I assume $x,y>0$.

If $x=y$, we arrive at $f(x)=x\ln x=\ln k$. Function $f$ decreases on $(0,1/e)$ from $1$ to $-1/e$, and then increases to infinity on $(1/e,+\infty)$. So there are two solutions of such type for $k\in(e^{-1/e},1)$, one for $k=e^{-1/e}$, and one for $k\geq 1$.

Now assume that $x<y$. Set $s=\ln x$, $t=\ln y$. We have $se^t=te^s=\ln k$, so that $g(s)=g(t)$, where $g(x)=xe^{-x}$. We have $g’(x)=e^{-x}(1-x)$, so $g$ increases on $(-\infty, 1)$ from $-\infty$ to $1/e$, and then decreases on $(1,+\infty)$ from $1/e$ to $0$. Hence we may assume that $s\in (0,1)$ and $t>1$.

Regarding $t$ as an implicit function of $s$ we are to investigate the function $h(s)=te^s$ on $s\in(0,1)$. We have $$ t’= \frac{e^{-s}(1-s)}{e^{-t}(1-t)} =\frac{t(1-s)}{s(1-t)}. $$ Hence $$ h’(s)=e^s\left(t+ \frac{t(1-s)}{s(1-t)}\right) =te^s\frac{1-st}{s(1-t)}. $$

Lemma. $st<1$.

Proof. Due to behavior of $g$, the claim is equivalent to $g(s)>g(1/s)$, or $2\ln s>s-1/s$. This inequality turns into equality at $s=1$, while the derivatives of both parts satisfy $2/s<1+1/s^2$. The claim follows.

Thus we have $h’(s)<0$ for all $s\in(0,1)$. So the range of $h$ is $(h(1), h(+0))=(e,+\infty)$. Hence such solution exists for $k>e^e$ and is unique for such $k$ (apart from permitting $x$ and $y$).