Logarithm is of base 2, please help to solve.

And does f₇(n) = 2^lognlogn ?

Question 1 [8 marks]: Rank the following functions in increasing order of growth; that is, the function f(n) is before function g(n) in your list, only if f(n)=O(g(n)). - f1(n)=nn/2 - f2(n)=(2n)n - f3(n)=(logn)logn - f4(n)=8logn - f5(n)=n4/3 - f6(n)=n3n2 - f7(n)=2(logn)2 - f8(n)=nlogn To simplify notations, we write f(n)