I need help proving that if f(n) = O(g(n)) implies 2^(f(n)) = O(2^g(n)))

Question

In a previous problem, I showed (hopefully correctly) that f(n) = O(g(n)) implies lg(f(n)) = O(lg(g(n))) with sufficient conditions (e.g., lg

Answer 1

If f(n) = O(g(n)),
2^(f(n)) not equal to O(2^g(n)))

Let, f(n) = 2log n and g(n) = log n
(Assume log is to the base 2)

We know, 2log n <= c(log n) therefore f(n) = O(g(n))

2^(f(n)) = 2^log n^2 = n^2
2^(g(n)) = 2^log n = n

We know that
n^2 is not O(n)

Therefore, 2^(f(n)) not equal to O(2^g(n)))