The given equation: $$\frac{8} {\ln(2)} \cdot \ln(x) = x.$$
Some algebra, and then we get the solutions:
$$x = e^{W_0(-\frac{\ln(2)} {8})}\approx 1.1 ,\;\; x = e^{W_{-1}(-\frac{\ln(2)} {8})} \approx 43.5593.$$
I found $W_{0}$ with Taylor series.
But there is addition real solution, $W_{-1}$.
How I find the $W_{-1}$?