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

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

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