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This question already has an answer here:

I want to prove: $$n^p>\log n$$ for a fixed $p\in (0,1)$ when $n\rightarrow \infty$. I test by some big number and it seems true. But how to prove that?

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marked as duplicate by Nosrati, rtybase, Adrian Keister, Simply Beautiful Art, Lord Shark the Unknown Aug 1 '18 at 14:36

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  • $\begingroup$ See this answer containing a proof of this result, to prove another ... $\endgroup$ – rtybase Aug 1 '18 at 12:04
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The hint:

Take $k$ such that $n=e^k$.

Now, let $e^p=1+t$.

Thus, $t>0$ and we need to prove that: $$(1+t)^k>k$$ for $k\rightarrow+\infty$.

I think it will help.

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Use the inequality $\log x<x-1$ so that if $q>0$ then $$\log n^q<n^q-1<n^q$$ or $$\log n<\frac{n^q} {q} $$ Now take $q=p/2$ and then we have $$\log n<2n^{p/2}/p$$ and this is less than $n^p$ if $n^p>4/p^2$ or if $n>(2/p)^{2/p}$.

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