# Solving $x^y = y^x$ analytically in terms of the Lambert $W$ function

I'm interested in deriving the solution for $y$ in terms of $x$ given $x^y = y^x$ using the Lambert $W$ function. Wolfram Alpha states:

$$y = - \frac{x\ W\left(-\frac{\log(x)}{x}\right)}{\log(x)}$$

So far I have done the following:

\begin{align*} x^y & = y^x\\ y \log(x) & = x \log(y)\\ \log(y)/y & = \log(x)/x\\ \log(y)/y & = \alpha && (\alpha=\log(x)/x) \end{align*}

The rest of it is proving the solution for $y$ in the last equation is $y = - W(-\alpha)/\alpha$. I can easily verify the solution but I'm unsure how to derive it.

Starting from $$\frac{\ln x}{x}=\frac{\ln y}{y}$$ let $y =\mathrm{e}^{-u}$ we find $$u\mathrm{e}^{u} = -\frac{\ln x}{x}$$ taking the lambert we find $$u = W\left(-\frac{\ln x}{x}\right)$$ remember $u=-\ln y = -\frac{y}{x}\ln x$ so we get $$-\frac{y}{x}\ln x = W\left(-\frac{\ln x}{x}\right) \implies y = \frac{-xW\left(-\frac{\ln x}{x}\right)}{\ln x}$$
• How can I get $y=-4$ from $x=-2$ here? – user286485 Apr 13 '18 at 16:38
You do not need to apply logarythm on both sides. $$x^y=y^x\\ e^{y\ln x}y^{-x}=1\\ e^{\frac{y\ln x}{-x}}y=1\\ e^{\frac{y\ln x}{-x}}\frac{y\ln x}{-x}=\frac{\ln x}{-x}\\ \frac{y\ln x}{-x}=W\left({\frac{\ln x}{-x}}\right)\\ y=\frac{-xW\left({\frac{\ln x}{-x}}\right)}{\ln x}$$