Explicit solutions to $x^y=y^x$ using Lambert $W$

Question

A Flammable Maths video gives the solutions to the title equation by $y=-\frac{x}{\ln x}W(-\frac{\ln x}{x})$. This makes a lot of sense, given that Wikipedia gives $W_0(-\frac{\ln x}{x})=-\ln(x)$ for $x<e$ and $W_{-1}(-\frac{\ln x}{x})=-\ln(x)$ for $x>e$, resulting in the trivial solutions on $y=x$. What I'm interested in is the following:

Is there a way to get the other solution, i.e. $y=-\frac{x}{\ln x}W_0(-\frac{\ln x}{x})$ for $x>e$ and $y=-\frac{x}{\ln x}W_{-1}(-\frac{\ln x}{x})$ for $x<e$, without a piecewise function?
Can this be simplified to a nice expression in a similar way to the $W$ $ln$ simplification of the trivial solution?

This isn't a direct answer to your question, but you might be interested in this post as well — Ben Grossmann, May 27 '20 at 16:32
That polar solution is really interesting, thanks! The other solutions come back to this function, though, and I'd still like to see if there's a simplification here :) — Erez Abrams, May 27 '20 at 16:36
I suspect that the answer to this question is something like the following: when extracting a solution for one variable in a relation like this, we always have to make some "ugly" separations or piecewise approaches. I think that an (easier) equation whose solution follows a similar process is $$ x-y = (x+y)^2 $$ — Ben Grossmann, May 27 '20 at 16:49
So there is no solution to the second point either? i.e., there is no way to simplify $W_0(-\frac{\ln x}{x})$ for $x>e$? — Erez Abrams, May 27 '20 at 17:03
Does this answer your question? Is there analytic solution to $x^y=y^x\land x\neq y$ as $y(x)$? — Jean Marie, May 27 '20 at 17:04
@JeanMarie it does not; I've already arrived to the same answer as that post, but am looking for a further simplification; after Omnomnomnom's point, I'm basically looking for a simplification of $W_0(-\frac{\ln x}{x})$ for $x>e$. — Erez Abrams, May 27 '20 at 17:10

score 1 · Answer 1 · answered May 31 '20 at 17:56

1

Another parametric solution in terms of $a>0$ is

\begin{align} x&=a^{1/(a-1)} ,\\ y&=a^{a/(a-1)} =a\,x . \end{align}

answered May 31 '20 at 17:56

g.kov

13,841

Explicit solutions to $x^y=y^x$ using Lambert $W$

1 Answers1