Pre-interview_Test_Question_22

Screenshot 2020-07-05 at 23.22.14

We have:

\log(f(n)) = 2^{2^n} \log 2

and:

\log(g(n)) = 100^n \log 100

Going again:

\log \log f(n) = \log(2^{2^n}) + \log \log 2 = 2^n \log 2 + \log \log 2

and:

\log \log g(n) = \log(100^n) + \log \log 100 = n \log 100 + \log \log 100

Note that since \displaystyle \frac n {2^n} \to 0 as n \to \infty (can use L’Hopital’s rule or similar), \log \log f(n) is eventually much larger than \log \log g(n), so f(n) is eventually much larger than g(n) since \log is increasing.

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