Supposing $f(N)\leq\frac{\ln\big(\frac{\ln N}{2\alpha\ln\ln N}\big)}{\ln2}$, $\alpha>0$ and $\epsilon>0$ holds then what is the largest $f(n)$ and smallest $\alpha,\epsilon$ we can choose such that $$(\ln N)^{\epsilon f(N)}2^{f(N)}(\ln N)^{\alpha(2^{f(N)}-1)}=\sqrt{N}$$ holds?
Can $f(N)=\frac{\ln\big(\frac{\ln N}{2\alpha\ln\ln N}\big)}{\ln2}-\epsilon$ hold at any small $\epsilon>0$ at some appropriate $\alpha>0$? How close can we get to this?