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Function 1: $2^{\log_*n}$

Function 2: $\log(\log n)$

The first function is 2 to the log-star of $n$, the second function is log of log of $n$. What I need to know is which one is Big-Omega of the other one, which means, which one grows faster. How can I figure that out? I know the definition of log-star (iterative logarithm) but I don't know how to apply it in order to compare the given functions. Could you guys give some help?

Thank you in advance.

David Richerby
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1 Answers1

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Consider $n_k=2^{2^{2^{\cdots}}}$, $k$ levels. Then $\log\log n_k = 2^{2^{2^{\cdots}}}$, $k-2$ levels, while $2^{\log^* n_k} = 2^k$. Does this help?

Yuval Filmus
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