# minimum value of n such that $f(x)>0$ $\forall x\geq n$

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.

Hint: No need for derivatives; $$f(x) >0$$ iff $$(\ln x -\frac 1 {2\ln 2})^{2} >1+\frac 1 {(2\ln 2)^{2}}$$. Can you finish?
• @mathisfun they completed a square with your function $f(x)$, and they're asking for $f(x)>0$. Square it back out and bring everything to the lhs to see. Jun 12 '20 at 11:24
• This is also giving $n=4$ Jun 12 '20 at 14:57