Let $f(n)$ and $g(n)$ be asymptotically positive functions.
Prove or disapprove the statement: $$f(n)=O(g(n)) \Rightarrow 2^{f(n)}=O(2^{g(n)})$$
That's what I have tried:
$$f(n)=O(g(n)), \text{ so } \exists c>0 \text{ and } n_0 \geq 1 \text{ such that } \forall n \geq n_0: f(n) \leq c g(n) \Rightarrow 2^{f(n)}\leq2^{cg(n)} (1)$$
$$2^{f(n)}=O(2^{g(n)}) \text{ means that }: \exists n_0' \geq 1 \text{ and } c'>0 \text{ such that } 2^{f(n)} \leq c'2^{g(n)} (2)$$
For $f(n)=cn, g(n)=n: (1) \Rightarrow 2^{cn} \leq 2^{cn} \checkmark$
$(2) \Rightarrow 2^{cn} \leq c' 2^{n} \Rightarrow 2^{(c-1)n} \leq c' \Rightarrow (c-1)n \leq lg c' \leq n \Rightarrow n \leq \frac{\lg c'}{c-1} \text{ that cannot be true,because it must be } n \geq n_0$
Could you tell me if it is right?
I think your argument is mostly correct - assuming you're trying to disprove the statement. But remember: disproving a universal statement means just exhibiting a single counterexample. I like your $f$ and $g$, but I wouldn't use the same $c$ in the definition of $f$ and in the $O$-property. So just show that $f(n)=2n$, $g(n)=n$ is a counterexample. Showing $f(n)=O(g(n))$ is easy of course. Then you have to show that $2^{2n}$ is not $O(2^n)$. In other words, you need to prove the negation of the $O$ definition, which is:
For every $c>0$ and for every $n_0\ge1$, there exists $n\ge n_0$ such that $2^{2n} > c2^n$.
Keeping track of exactly what statement you have to prove is half the battle (that's what will convert your ideas and preliminary computations into a proof).