# Prove that for sufficiently large $x$, $(\log x)^p<x$ where $p$ is a positive integer. [duplicate]

Ok so I was playing with GeoGebra by plotting various functions, then I come up with the following kind of graph when I plot $$\log(x)$$ to various positive powers and $$f(x)=x$$.

The green one is the function $$f(x)=x$$ and the blue one is $$g(x)=(\log(x))^4$$. So it seems for sufficiently large $$x$$, any positive power of $$\log(x)$$ is less than $$x$$. But I am unable to prove this fact. Can anyone give me a hint as how to approach to this problem?

## marked as duplicate by rtybase, Nosrati, José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 23 '18 at 19:00

• Is $e^y>y^p$ for sufficiently large $y$? – Lord Shark the Unknown Sep 23 '18 at 5:34
• repeated application of L'Hopital's rule – user254433 Sep 23 '18 at 5:51
• As I have mentioned often here, one just needs to pick a number $q$ with $0<q<1/p$ (one choice is $q=1/(p+1)$) and the use the inequality $\log t\leq t-1<t$ with $t=x^q$ and get $\log x<x^q/q$. Since $q<1/p$ the conclusion easily follows. – Paramanand Singh Sep 23 '18 at 15:10

We have $$\log^p(x)<x\iff\log(x)<x^{1/p}\iff\frac{\log(x)}{x^{1/p}}<1$$

Taking the limit to check the behaviour when $x$ is large: $$\lim_{x\to\infty}\frac{\log(x)}{x^{1/p}}=\lim_{x\to\infty}\frac{\frac{d}{dx}\log(x)}{\frac{d}{dx}x^{1/p}}=\lim_{x\to\infty}\frac{1/x}{x^{1/p-1}/p}\\=\lim_{x\to\infty}\frac{p}{x^{1/p}}=0<1$$

So for sufficiently large $x$ we have $\log^p(x)<x$

Clearly this is true iff $$x < e^{x^{1/p}}$$ for sufficiently large x (or $$x^p < e^x)$$.

What if we proved that for sufficiently large $$x$$, $$m$$, $$(1+ 1/m)^{mx} > x^p$$? Also, derivatives have some interesting properties.

To prove that $$0=\lim_{x\to \infty}(\log x)^Ax^{-B}$$ for any $$A,B>0$$ by elementary means, first observe that $$0=\lim_{x\to \infty}(\log x)^Ax^{-B}\iff$$ $$\iff 0=\lim_{x\to \infty}((\log x)x^{-B/A})^A\iff$$ $$\iff 0=\lim_{x\to \infty} (\log x)x^{-B/A} .$$ For brevity let $$y=(B\log x)/A.$$ Let $$[y]$$ denote the largest integer not exceeding $$y.$$ If $$x>e^{2A/B}$$ then $$y\geq [y]\geq 2.$$ So by the Binomial Theorem, if $$x>e^{2A/B}$$ then $$x^{B/A}=e^y>2^y\geq 2^{[y]}=\sum_{j=0}^{[y]}\binom {[y]}{j}>$$ $$> \binom {[y]}{2}=[y]([y]-1)/2>$$ $$>([y]-1)^2/2>(y-2)^2/2.$$ Now $$x>e^{2A/B}$$ also implies $$x^{B/A}>2$$. Therefore if $$x>e^{2A/B}$$ then $$(A/B)\sqrt {3x^{B/A}}\;>(A/B) \sqrt {2x^{B/A}+2}\;>$$ $$>(A/B)y=\log x$$ and hence $$(A/B)(\sqrt 3\;) x^{-B/2A}>(\log x)x^{-B/A}>0.$$ So $$\lim_{x\to \infty}(\log x)x^{-B/A}=0.$$

Note that, with the substitution $$t=x^{1/p}$$, $$\lim_{x\to\infty}\frac{x^{1/p}}{\log x}= \lim_{t\to\infty}\frac{t}{p\log t}=\infty$$ with a very simple application of l'Hôpital.