# Comparing asymptotic order of logarithmic functions

If I have two complicated logarithmic functions, say $\sqrt{\log n}$ and $\log(n(\log n)^3)$, and I have to compare them in terms of their asymptotic order. How do I do that? Do I have to create graphs, or is there another definitive way of doing that?

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## 1 Answer

Using elementary properties of the logarithm we find out that $$\log(n \log^3 n) = \log n + 3\log\log n = O(\log n),$$ whereas $\sqrt{\log n} = o(\log n)$. So $\sqrt{\log n}$ grows more slowly.

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OK, got it just one more question,how do I compare log(n) and log(log n) then? –  Parth Mody Sep 21 '12 at 0:56
From $\log n = o(n)$ it follows that $\log\log m = o(\log m)$ (substitute $n = \log m$). That $\log n = o(n)$ can be shown using l'Hopital's rule. –  Yuval Filmus Sep 21 '12 at 2:11
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