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

Well, it's not even true to begin with.

Let's say algorithm A takes 2n steps, and algorithm B takes n steps. Then their ratio is a constant.

But the ratio of 2²ⁿ and 2ⁿ is not a constant, so what you said doesn't hold.