# Comparing order of growth with n^(logn) - need to know where value comes from?

I'm trying to understand how you compare the following for order of growth.

With the below working out with $$f_4$$ I don't get where the $$x^5$$ and $$x^6$$ come from at the end. With $$f_4$$ you have $$n^{\log n}$$ when for example $$\log 64$$ doesn't equal 6? Can someone please explain where $$x^5$$ and $$x^6$$ come from?

Working out:

• $$n = 32$$, $$f_1 = 2^{32}$$, $$f_4 = 32^5 = 2^{25}$$
• $$n = 64$$, $$f_1 = 2^{64}$$, $$f_4 = 64^6 = 2^{36}$$

Compare these below

• $$f_1(n) = 2^n$$
• $$f_2(n) = n^{3/2}$$
• $$f_3(n) = n \log n$$
• $$f_4(n) = n^{\log n}$$
• (Think dualis instead of naturalis.) Dec 20, 2021 at 10:45
• @greybeard how do you compare order of growth with (2^20) * n i.e how would you work that out? Dec 20, 2021 at 11:19
• Well actually, $\log_2 64$ does equal $6$. Dec 20, 2021 at 12:39

When you want to evaluate $$f_4 = n^{\log_2 n}$$ for some number, let's say 256, you first evaluate the exponent, $$\log_2 n = \log_2 256 = 8$$, then you put it into the full formula as $$n^8 = 256^8$$. Of course, since $$256^8 = \left( 2^{ 8 } \right)^8 = 2 ^ {8 \cdot 8 } = 2^ {64}$$.
Let's now do the exercises you have: for $$n=32$$, we have that $$\log_2 n = \log_2 32 = 5$$, hence $$n^{5} = 32 ^ 5 = \left(2^{5}\right)^5 = 2^{5 \cdot 5} = 2^{25}$$.
For $$n = 64$$, we have that $$\log_2 n = \log_2 64 = 6$$, thus $$n^{6} = 64 ^ 6 = \left(2^{6}\right)^6 = 2^{6 \cdot 6} = 2^{36}$$.
It seems to me that $$n^{\log_2 n}$$ follows the pattern of $$2^{x^2}$$ where $$x = \log_2 n$$.