Simplifying logarithm question

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

Answer 1

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 $n=2^4= 16$.

Answer 2

\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}