TADM2E 2.19
From Algorithm Wiki
$ ({1}/{3})^n < 6 < \log \log n < \log n = \ln n < (\log n)^2 < n^{\frac{1}{3}} + \log n < \sqrt{n} < \frac{n}{\log n}< n < n \log n < n^2 = n^2 + \log n < n^3 < n - n^3 + 7n^5 < ({3}/{2})^n < 2^n < n! $
$ ({1}/{3})^n < 6 < \log \log n < \log n = \ln n < (\log n)^2 < n^{\frac{1}{3}} + \log n < \sqrt{n} < \frac{n}{\log n}< n < n \log n < n^2 = n^2 + \log n < n^3 < n - n^3 + 7n^5 < ({3}/{2})^n < 2^n < n! $
