# How to prove $n^p>\log n$ for a fixed $p\in (0,1)$ when $n\rightarrow \infty$? [duplicate]

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?

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

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

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 $k\rightarrow+\infty$.
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 $n>(2/p)^{2/p}$.