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To do the question you basically have to divide f(n) by first 2^n and see if the limit as n approaches infinity of f(n) over g(n) which is 2^n = 0, infinity, or a constant. If its 0 you find the next limit which is n^k and see if it equals 0, infinity, or a constant. If its infinity, then f(n) = big omega of that function. If its 0 you go on to the next one and so on Then n^a, then logn then a constant. The base for the log are 2 and not 10, so if you do the log then it will simplify 2^n to just n.

Discuss the growth of the below functions (Show the work) f(n)= (logn) bog b. f(n) = 2V7691 6. f(n)=(1236 d. f(n) = Kan