How to solve: $\frac{x}{\log_2(x) }= y$

Question

For example, I can solve: $x \log_2(x) = y$

$x \log_2(x) = x \log_e(x) / \log_e(2) = e^{\log_e(x)} \log_e(x) / \log_e(2)$

$e^{\log_e(x)} \log_e(x) = y\log_e(2)$

$e^{W(z)} W(z) = z$, where W(z) is the Lambert W-function

$log_e(x) = W(y\log_e(2))$

$x = e^{W(y\log_e(2))}$

But how to (find $x$) solve: $\frac{x}{\log_2(x)} = y$

Answer:

$$\displaystyle\begin{array}$x&=& \frac{1}{e^{W(ln(\frac{1}{2})^{1/y})}} = \frac{1}{e^{W(−ln(2)/y)}} \end{array}$$

Answer 1

$$\begin{array}{rcl} \displaystyle\frac{x}{\log_2(x)}&=&y\\ \displaystyle\frac{x}{\ln(x)}&=&\displaystyle\frac{y}{\ln(2)}\\ \displaystyle\frac{\ln(x)}{x}&=&\displaystyle\frac{\ln(2)}{y}\\ \displaystyle\frac{-\ln(x)}{x}&=&\displaystyle-\frac{\ln(2)}{y}\\ \displaystyle\frac{1}{x}{\ln(\frac{1}{x})}&=&\displaystyle-\frac{1}{y}\ln(2)\\ \end{array}$$ Now, from Wikipedia, $W(t\ln(t))=\ln(t)$ so $$\begin{array}{rcl} \displaystyle W\biggl(\frac{1}{x}\ln\left(\frac{1}{x}\right)\biggr)&=&\displaystyle W\biggl(-\frac{1}{y}\ln(2)\biggr)\\ \displaystyle\ln\left(\frac{1}{x}\right)&=&\displaystyle W\biggl(-\frac{1}{y}\ln(2)\biggr)\\ \displaystyle\ln\left(x\right)&=&\displaystyle -W\biggl(-\frac{1}{y}\ln(2)\biggr)\\ \Rightarrow x&=&\displaystyle\mathrm{e}^{-W\biggl(-\frac{1}{y}\ln(2)\biggr)} \end{array}$$

Answer 2

Maple gives

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