Let, $ f(x)= (ln x)^2 -1- \frac{ln x}{ln 2}$ .Then what is the minimum value of $n \in \mathbb{N}$ such that $f(x)>0$ $ \forall x\geq n$?
I tried finding $f'(x)$ but I am getting $n=4$ which is not correct. In fact, my calculations suggest $n=8$? But how can I prove it?
Any help would be appreciated. Thanks in advance.