# Which is greater: $n^{1.01}$ or $n\cdot log_{10}(n)$ ?

Can someone please explain how the right side can be less than the left side? I have plugged numerous numbers into n and every time I get the left side being less than the right side. My professor is convinced the right side is less than the left side. He has a PHD in math so he should be right. I just don't understand his explanation.

$n^{1.01} < n\cdot log_{10}(n)$

$1000^{1.01} < 1000*log_{10}(1000)$

$1071.51 < 3000$

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The question amounts to asking which is larger, $\sqrt[100]{n}$ or $\log(n)$. For sufficiently large $n$, what do you think? –  Jonathan Y. Nov 8 '13 at 2:32

In the race between a (positive) power and a logarithm, the power wins eventually. So $n^{0.01} > \lg(n)$ for all sufficiently large $n$, and thus $n^{1.01} > n \lg(n)$ for those same $n$. But how large is "sufficiently large"? In this case, $n > 3.8125 \times 10^{237}$ approximately.
@Sush Yes. The usual method is to use L'Hospital's rule. You can use it to show that $\ln^a(x)/x^b$ converges to $0$ as $x$ diverges to $\infty$ for any $a,b>0$. So any power of $x$, no matter how small, eventually grows (much) larger (and faster) than any power of $\ln$, no matter how large. As Robert's posts indicates, "eventually" can be a very long time. –  zibadawa timmy Nov 8 '13 at 3:11
@zibadawatimmy, Thanks a lot. I think it should be $\log_a(x)/x^b$, right? –  Sush Nov 8 '13 at 3:15
How were you able to calculate $3.8125 \times 10^{237}$ as being "sufficiently large" enough? That number is so big I would have no idea where to start. –  cokedude Nov 8 '13 at 3:44
I think we pupils should use Wolfram|Alpha. Note that we get two solutions for $n$, but at $n=2.746\dots$ , graph of $log (n)$ crosses graph of $n^{0.01}$ from beneath and thus after $n=3.8125 \times 10^{237}$ only, we get that $n^{1.01} > n \log(n)$. Note also that, for interval $0\leq n< 2.746\dots$,too, inequality $n^{1.01} > n \log(n)$ holds. –  Sush Nov 8 '13 at 5:34