Explicit components of $y^x=x^y$

Question

When graphing the implicit equation $y^x=x^y$ it seems like it consists of two possibly explicit curves. On the one hand there is the linear solution $y=x$. But as it appears there is a second curve monotonously decreasing as x increases, intersecting with the curve $y=x$ at coordinate point $(2,2)$. Is there any way to find out the explicit form of this other curve?

Answer 1

You can get a parametric set of equations. Define $y=ax$ Then $$x^y=y^x\\x^xa^x=x^{ax}\\a=x^{a-1}\\x=a^{\frac 1{a-1}}\\y=a^{\frac a{a-1}}$$

Answer 2

Yes, it's possible, under certain definitions of the word "possible".

First of all, it's a clearly well-defined concept so there is a mathematical function that for any given $x>1$ gives you the $y>1$ such that

• If $x = e$, then $y = e$
• If $x \neq e$, then $y\neq x$
• $x^y = y^x$

However, it is not easily described. Plugging it into WolframAlpha, we get the result $$ y = -\frac{x\,W\left(-\frac{\ln(x)}{x}\right)}{\ln(x)} $$ where $W$ is the Lambert $W$ function (the inverse of $x\mapsto xe^x$), and personally I have no trouble believing that this is the simplest desciption. I'm sure it's possible to reverse-engineer a proof of this relation now that we have the answer, but I do not envy anyone who would have to do this blind.

Answer 3

If $x^y = y^x$, then taking logs of both sides gives $y \log x = x \log y$, so $\frac{\log x}{x} = \frac{\log y}{y}$. The function $f(x) = \frac{\log x}{x}$ satisfies $f(1) = \lim_{x \to \infty} f(x) = 0$, and it is increasing up to $x = e$ and decreasing afterwards. This tells us that for $x > 1$, there are exactly two values of $y$ such that $f(y) = f(x)$, including $y = x$ and one value of $y$ on the other side of $e$. (Except of course $x = e$, for which only $y = e$ works.) In particular, this implies that the two curves intersect at $(e, e)$, not $(2, 2)$.

Unfortunately it seems like the inverse of $f$ (say, the one producing values between $1$ and $e$, or the one between $e$ and $\infty$) isn't an elementary function. As Arthur pointed out in his answer, WolframAlpha expresses the result in terms of the Lambert W function, which is related to the inverse of $f$.

Answer 4

Arthur gave you an explicit form using the Lamber W-function. I present you a more algorithmic approach using power towers. Rearrange $x^y=y^x$ to $$x=\sqrt[y]{y}^x.$$ So when you define $z:=\sqrt[y]y$, then for a given $y$ you can compute $$x=z^{z^{z^{\cdots}}}.$$ It is not obivous that this even converges, but check by yourself by using the recursice rule $x_0=1$ and $x_{n+1}=z^{x_n}$.