# Compare growth rate of functions

I was given homework to sort some (14) functions in order of their growth rate. I am confused about two functions $3^\sqrt{\log n}$ and $n^{\log n}$: about where these two lie within those 14 functions.

I tried wolframalpha but it does not plot the graphs well and is not very useful. Which technique I should use to compare growth rates of functions?

EDIT:
I tried taking logs, but it also confused at places. For example I have taken double logs below

log(5) + log(log(n)) AND log(log(5n+20))

I don't know which one is bigger here

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Hint: take log's. – David Speyer Nov 27 '11 at 18:35
@DavidSpeyer I tried log, but it also confused a bit, see the example above – LifeH2O Nov 27 '11 at 18:40

You could just compare the two functions. The base 3 is constant and the base $n$ tends to infinity. Now compare the exponents: $\sqrt{\log n}$ grows more slowly than $\log n$. These two observations imply that $3^{\sqrt{\log n}}$ grows more slowly that $n^{\log n}$.
In fact, for $n>10$: $${ n^{\log n}\over3^{\sqrt{\log n}}}\ge { 10^{\log n}\over3^{\sqrt{\log n}}} \ge{ 10^{\log n}\over3^{ \log n }}={(10/3)^{\log n}}\rightarrow\infty.$$
There, you need to compare the growth rates of $n$ and $5n+20$. They grow at the same rate since ${n\over 5n+20}={1\over 5+{20\over n}}\rightarrow {1\over 5}$. Since these grow at the same rate, so will their logarithms (and adding $\log 5$ to one won't change that fact). – David Mitra Nov 27 '11 at 19:14
Although the growth rates are the same, $\log \log(5 n + 20) < \log (5) + \log \log(n)$ for large $n$ (in fact for $n \ge 2$). – Robert Israel Nov 27 '11 at 19:51
You meant $\log (5)+\log(\log(n))$ and $\log(\log(n^5+20))$? Then $\log(\log(n^5+20))$ grows as fast as $\log(\log(n^5))= \log (5\log n)=\log 5 +\log(\log n)$. – David Mitra Nov 27 '11 at 19:56