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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.

Thanks in advance.

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2 Answers 2

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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 $$ -\frac{y}{x}\ln x = W\left(-\frac{\ln x}{x}\right) \implies y = \frac{-xW\left(-\frac{\ln x}{x}\right)}{\ln x} $$

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  • $\begingroup$ This is awesome. Exactly what I was looking for. Thanks. $\endgroup$ May 21, 2015 at 15:28
  • $\begingroup$ How can I get $y=-4$ from $x=-2$ here? $\endgroup$
    – user286485
    Apr 13, 2018 at 16:38
  • $\begingroup$ @user286485 You can't, because this solution is only about positive numbers $\endgroup$ Jul 24, 2018 at 20:07
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You do not need to apply logarythm on both sides. $$ x^y=y^x\\ e^{y\ln x}y^{-x}=1\\ e^{\frac{y\ln x}{-x}}y=1\\ e^{\frac{y\ln x}{-x}}\frac{y\ln x}{-x}=\frac{\ln x}{-x}\\ \frac{y\ln x}{-x}=W\left({\frac{\ln x}{-x}}\right)\\ y=\frac{-xW\left({\frac{\ln x}{-x}}\right)}{\ln x} $$

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