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?

  • $\begingroup$ Can you explain little more? $\endgroup$ Jun 12 '20 at 11:14
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.