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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.

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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?

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  • $\begingroup$ Can you explain little more? $\endgroup$ Jun 12 '20 at 11:14
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    $\begingroup$ @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. $\endgroup$
    – snulty
    Jun 12 '20 at 11:24
  • $\begingroup$ Are you talking about square root? That is I need to take square root both side? $\endgroup$ Jun 12 '20 at 13:42
  • $\begingroup$ This is also giving $n=4$ $\endgroup$ Jun 12 '20 at 14:57

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