Solving for $x$ an $x\log(x)$ kind of equation Given the following:
$$\frac{x \log\left(\frac{4x}{d}\right)}{y_0} = 10$$
where $d$ and $y_0$ are constants, how can I solve for $x$?
 A: Hint
Use Lamber-W function. It is a function $W(x)$ for which
$$e^{W(x)}W(x)=x$$
Apply $\exp$ to both sides of your original equation
$$e^x\cdot\frac{4x}{d}=e^{10y_0}$$
Do you recognize where you can apply $W(x)$?
A: For the equation
$$\frac{x \log\left(\frac{4x}{d}\right)}{y_0} = 10$$
multiply both sides by $4 y_{0}/d$ to obtain
$$\frac{4 x}{d} \, \log\left(\frac{4 x}{d}\right) = \frac{40 y_{0}}{d}.$$
Now using the Lambert W-function defined by $x e^{x} = W(x)$ with the property $e^{W(x)} = \frac{x}{W(x)}$ then
\begin{align}
\frac{4 x}{d} \, \log\left(\frac{4 x}{d}\right) &= \frac{40 y_{0}}{d} \\
t \, \log(t) &= \frac{40 y_{0}}{d} \hspace{15mm} t = \frac{4 x}{d} \\
u \, e^{u} &= \frac{40 y_{0}}{d} \hspace{15mm} u = \log(t) = \log\left(\frac{4 x}{d}\right) \\
u &= W\left(\frac{40 y_{0}}{d}\right) \\
\log\left(\frac{4 x}{d}\right) &= W\left(\frac{40 y_{0}}{d}\right) \\
\frac{4 x}{d} &= e^{W\left(\frac{40 y_{0}}{d}\right)} \\
x &= \frac{\frac{d}{4} \, \frac{40 y_{0}}{d} }{W\left(\frac{40 y_{0}}{d}\right)} = \frac{10 \, y_{0}}{W\left(\frac{40 y_{0}}{d}\right)}.  
\end{align}
