Solutions of $x^{y} = y^{x}$

Asked May 18 '21 at 18:27

Active May 18 '21 at 19:08

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I am attempting to find solutions to $x^y = y^x$.

I have been trying to find solutions to this equation for quite some time now, and using all the math (and common sense) I know, I arrived at one possible non trivial solution, $x = 4, \;y =2.$

Are there other solutions to this equation and if there are how do I find them?

edited May 18 '21 at 19:08

amWhy

209,954

asked May 18 '21 at 18:27

iAmSecretlyFlash

So $x^y=x$? Is that your equation? The solutions are x,y=1 and some ugly y=imaginary answers with ln(x) – Тyma Gaidash May 18 '21 at 18:29
It’s $x^y=y^x$. – iAmSecretlyFlash May 18 '21 at 18:30
Blackpenredpen solved this: video – Тyma Gaidash May 18 '21 at 18:31
Please use MathJax. Here's a tutorial – Arturo Magidin May 18 '21 at 18:32
This article might be useful – Fabio Ori May 18 '21 at 18:33
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Does this answer your question? $x^y = y^x$ for integers $x$ and $y$ Though the question asks about integer solutions, the accepted answer provides the general solution. – Arturo Magidin May 18 '21 at 18:33
Nice find, @Arturo! – amWhy May 18 '21 at 19:09
Using the Lambert W function you can get a general solution as $$ y x^{-\frac{y}{x}} =1 \implies -\frac{\color{blue}{y}}{x} \ln(x) \color{blue}{e^{-\frac{y}{x} \ln(x) } }= -\frac{\ln(x)}{x} \implies \boxed{y = -\frac{x}{\ln(x)} W\left(-\frac{\ln(x)}{x} \right)} $$ which can even be extended to obtain complex solutions. – Robert Lee May 18 '21 at 19:15
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In the next comment, I'll post a live Sage / Python script that solves for $m$ in $x=(1 + 1/m)^{m+1}$, without using the Lambert W. It's optimized for $x>e$, but it will work for $x>1.2$ or so. It uses Newton's method, and gets its initial approximation via $e\approx(1+1/m)^{m+1/2}$. (The Sage function n() just gets a numeric approximation of its arg; Sage works with symbolic expressions, when it can). – PM 2Ring May 18 '21 at 19:58
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$x^y=y^x$ script – PM 2Ring May 18 '21 at 20:03

Solutions of $x^{y} = y^{x}$

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