I'm looking for the minimum of the function ($n$, $o$, and $s$ are all constants greater than $0$):
$$f(x)=\sqrt{\frac{1}{4}n^{2}o^{2}*\left(n*x+x-\frac{2*ln(x)}{ln(2)}\right)^{2}+\frac{n^{2}s^{2}*ln(x)^{2}}{ln(2)^{4}}}$$
I can find a numerical answer for a given $n$, $s$, and $o$, but not an analytic solution.
I've tried taking the derivative with respect to $x$, which is:
$$\frac{\frac{n^{2}o^{2}\left(-\frac{2}{ln\left(2\right)*x}+n+1\right)\left(-\frac{2ln\left(x\right)}{ln\left(2\right)}+nx+x\right)}{2}+\frac{2n^{2}s^{2}ln\left(x\right)}{ln\left(2\right)^{4}*x}}{2\sqrt{\frac{n^{2}o^{2}\left(-\frac{2ln\left(x\right)}{ln\left(2\right)}+nx+x\right)^{2}}{4}+\frac{n^{2}s^{2}ln\left(x\right)^{2}}{ln\left(2\right)^{4}}}}$$
and then tried to solve for:
$$\frac{n^{2}o^{2}\left(-\frac{2}{ln\left(2\right)*x}+n+1\right)\left(-\frac{2ln\left(x\right)}{ln\left(2\right)}+nx+x\right)}{2}+\frac{2n^{2}s^{2}ln\left(x\right)}{ln\left(2\right)^{4}*x}=0$$
Again I can find a numerical solution for a given $n$, $s$, and $o$ but I can't find an analytic solution for $x$.
Does an analytic solution even exist, and if it does what is it/how would I find it?