When does $8n\log_2(n) = 2n^2$ What is the most systematic way to do this problem? I used the definition of logarithms and brute force to find $n = 16$, but I feel as though that was the worst way possible.
 A: After dividing by $n$ (and thereby removing the solution $n=0$) and simplifying, we get
$$
-\log(n)e^{-\log(n)}=-\frac{\log(2)}4
$$
Therefore,
$$
\begin{align}
n
&=e^{-\operatorname{W}\left(-\frac{\log(2)}4\right)}\\[6pt]
&=-\frac4{\log(2)}\operatorname{W}\left(-\frac{\log(2)}4\right)
\end{align}
$$
For negative arguments, Lambert W has two real branches giving two solutions: $1.2396277295227621418$ and $16$.
A: write your equation in the form
$$n\left(4\frac{\ln(n)}{\ln(2)}-n\right)=0$$
you can choose a numerical way or the LambertW function.
$$n_1=-4\,{\frac {{\rm W} \left(-1/4\,\ln  \left( 2 \right) \right)}{\ln 
 \left( 2 \right) }}
$$
$$n_2=-4\,{\frac {{\rm W} \left(-1,-1/4\,\ln  \left( 2 \right) \right)}{\ln 
 \left( 2 \right) }}
$$
$$n=0$$ is impossible
A: First of all, I would think one could also call $n=0$ a solution, because of $\lim_{n\to0}n\log_2(n)=0$. But discarding this solution and recalling that the question was whether we can do without brute force, let me suggest to divide the equation by $8n$ which gives $\log_2(n)=n/4$.  Antilog leads to $n=2^{n/4}$. Raising to the power 16 finally leads to $n^{16}=16^n$. This makes it quite obvious that $n=16$ is a solution, albeit not the only one.
