Closed form minimization of this function

Question

I'm trying to find a value of $x$ in terms of positive integers $n,b$ that minimizes this function: $$ \frac{b\,(n+2^x)}{x} \quad (x>0) $$ So far, I've tried differentiaton, but that doesn't lend itself to a clean solution. Any thoughts?

Answer 1

The zero of the derivative, which gives you the absolute minimum for the function when $x>0$ is in terms of the Lambert function, so there is not a "clean" solution. You'll need software to compute it

Answer 2

$$f(x)=\frac{b\,(n+2^x)}{x}\quad \implies \quad f'(x)=\frac b {x^2}\left(2^x (x \log (2)-1)-n\right)$$ the first derivative cancels when $$2^x (x \log (2)-1)-n=0\quad \implies \quad x_*=\frac{W\left(\frac{n}{e}\right)+1}{\log (2)}$$ where $W(.)$ is Lambert function. Checking $$f''(x_*)=\frac{b \,n \log ^3(2)}{W\left(\frac{n}{e}\right) \left(1+W\left(\frac{n}{e}\right)\right)} >0$$ and $$f(x_*)=\frac{b\, n \log (2)}{W\left(\frac{n}{e}\right)}$$

Let $n=t\,e$ to have a nicer form $$f_{\text{min}}= [b\,e\,\log(2)]\,\, e^{W(t)}$$ Look here at the plot of $ e^{W(t)}$