I am trying to solve:
$$8n^2=64 n\log(n)$$
I factored out $8n$ and was left with
$$n = 8 \log n$$
but I do not know how to proceed from here, I checked the solution on Wolfram Alpha, and they were able to simplify it down to:
which is approximately:
how where they able to get to this point ?
Best Answer
Your equation does not have a closed form in terms of standard functions you come across in school, but it does have a solution in terms of the Lambert W-function $W(z)$. The defining equation for $W(z)$ is $$z=W(z)e^{W(z)}$$ We want to rearrange your equation into something similar so we get $$\begin{align*}n&=8\log n \\ \longrightarrow\hspace{10mm}\frac n8&=\log n \\ \longrightarrow\hspace{7mm} e^{\frac n8}&=n \\ \longrightarrow\hspace{12mm} 1&=ne^{-\frac n8} \\ \longrightarrow\hspace{5mm} -\frac18&=-\frac n8e^{-\frac n8} \\\end{align*}$$ Now $$Y=Xe^X\iff X=W(Y)$$ Thus we get $$\begin{align*}-\frac n8&=W\left(-\frac18\right)\\ n&=-8W\left(-\frac18\right)\end{align*}$$ To get decimal approximations, you can numerically evaluate $W$ for its different branches and you get the solutions wolfram alpha gave you.