Simplifying logarithm question Without worrying about the background, I have a question that asks to solve for n. Pardon my formatting, but it seems understandable this way for the time being until I edit it:
$$4n^2 = 256 \log_2n$$
I'm not looking for the answer (it's 16), but I want to know the process to begin simplifying the right side.
Thank you in advance
 A: One way to solve it :
Since $n$ is an integer, and you have a base-2 $log$, then $n$ can be written as : $$n = 2^{k}$$ with $k > 0$ (at first).
After rearranging the initial expression, you end up with : 
$$k = 2^{2k-6}$$
Now, you can see that :
$k > 3$ and $k$ is even. So, if you try with $k=4$... $$2^{2*4 -6} = 2^{8-6} = 2^2 = 4 = k$$
And since $n=2^k$, you have $n=2^4= 16$.
A: 
\begin{align} 4n^2 &= 256 \log_2n \end{align}

This kind of equations can be solved in terms of the Lambert W function,
which provides a solution to equation of the form $x\exp(x)=y$
as $x=\operatorname{W}(y)$, hence we need 
to rearrange the terms in original equation 
in order to group terms $x$ and $\exp(x)$ together:
\begin{align}
n^2 &= 64 \frac{\ln(n)}{\ln(2)} \\
\ln(n) n^{-2} &=  \frac{\ln(2)}{64}\\
-2\ln(n) n^{-2} &=  -\frac{\ln(2)}{32}\\
\ln(n^{-2}) n^{-2} &=  -\frac{\ln(2)}{32}\\
\ln(n^{-2}) \exp(\ln(n^{-2})) &=  -\frac{\ln(2)}{32}\\
\ln(n^{-2}) &= \operatorname{W}\left(-\frac{\ln(2)}{32}\right) \\
n^{-2} &= \exp\left(\operatorname{W}\left(-\frac{\ln(2)}{32}\right)\right) \\
n &=\exp\left(-\frac{1}{2}\operatorname{W}\left(-\frac{\ln(2)}{32}\right)\right) \\
\end{align}
Since $-\mathrm{e}^{-1}<-\frac{\ln(2)}{32}<0$, 
there are two real solutions, corresponding 
to the two real branches of the Lambert W function, 
$\operatorname{W}_0$ and $\operatorname{W}_{-1}$:
\begin{align}
n &=\exp\left(-\frac{1}{2}\operatorname{W}_{0}\left(-\frac{\ln(2)}{32}\right)\right) 
\approx 1.01113448176 \quad (1)\\
n &=\exp\left(-\frac{1}{2}\operatorname{W}_{-1}\left(-\frac{\ln(2)}{32}\right)\right)
=16 \quad (2).
\end{align}
In case if the solution has to be an integer, 
(1) has to be ignored.
But the second solution is exact and integer,
since 
\begin{align}
\operatorname{W}_{-1}\left(-\frac{\ln(2)}{32}\right)
&=
\operatorname{W}_{-1}\left(-\frac{1}{128}\ln(16)\right)
\\
&=
\operatorname{W}_{-1}\left(
-2\ln(16)\exp(-2\ln(16))
\right)
=-2\ln(16).
\end{align}
