Suppose that $f (x)$ is $O(g(x))$. Does it follow that $2^{f(x)}$ is $O(2^{g(x)})$?

Question

Suppose that $f(x)$ is $O(g(x))$. Does it follow that ?

First, I start from for some $c$ is a real number. Then, I find . But, if i start from , I just find . I confused with that different form.

Answer 1

Consider $f(x) = x$ and $g(x) = x/2$ for $x\geq 0$. Then, $f(x) = 2g(x) = O(g(x))$, but you don't have $2^x = O(2^{x/2})$ (if $2^x \leq c 2^{x/2}$, then $x \leq \log(c) +x/2$, which is absurd). The answer to your question is no.