How to show that the following equality is true with the help of lambert W function?

Question

How to show that if $$y=x(2^{1/x}-1)\tag{1}$$ then $$x=-\frac{y\log(2)}{yW(-\frac{2^{-1/y}\log(2)}{y})+\log(2)}.\tag{2}$$ Where $\log$ is the natural logarithm and $W$ represent the lambert W function. Any help in this regard will be much appreciated. Thanks in advance.

Answer 1

\begin{eqnarray} y &=& x(2^{1/x} - 1) \\ x 2 ^{1/x} &=& y + x \\ \frac{2^{1/x}}{y} &=& \frac{1}{x} + \frac{1}{y} \\ \frac{1}{y} &=& \left(\frac{1}{x} + \frac{1}{y}\right)2^{-1/x} \\ \frac{\color{blue}{2^{-1/y}}}{y} &=& \left(\frac{1}{x} + \frac{1}{y}\right)2^{-1/x}\color{blue}{2^{-1/y}} \\ \frac{2^{-1/y}}{y} &=& \left(\frac{1}{x} + \frac{1}{y}\right)e^{-\ln 2/x}e^{-\ln 2/y} \\ \color{red}{-\ln 2}\frac{2^{-1/y}}{y} &=& \left(-\frac{\color{red}{\ln 2}}{x} - \frac{\color{red}{\ln 2}}{y}\right)e^{-\ln 2/x}e^{-\ln 2/y} \\ -\frac{2^{-1/y}\ln2}{y}&=& \left(\color{orange}{-\frac{\ln2}{x} -\frac{\ln2}{y}}\right)e^{\color{orange}{-\ln 2/ x - \ln 2 /y}} \\ \color{orange}{W}\left(-\frac{2^{-1/y}\ln2}{y} \right)&=& \color{orange}{-\frac{\ln2}{x} -\frac{\ln2}{y}} \\ \ln2 + y W\left(-\frac{2^{-1/y}\ln2}{y} \right) &=& -\frac{y}{x}\ln2 \\ \end{eqnarray}

And from here

$$ x = - \frac{y\ln 2}{\ln 2 + y W(-2^{-1/y} \ln 2/y)} $$