question about functions (asymptotic)

Question

This is right?

$f=\Omega(g)\Rightarrow2^f=\Omega(2^g)$?

If not I'd like to get a Counter-example.

Thank you!

Answer 1

No.

Let $f(x) = x$ and $g(x) = 2x$. Then $f(x) = \Omega(g(x))$ since $f(x) = \frac{1}{2} g(x)$ but for any $c>0$ it will not be true that $2^{f(x)} = 2^x \geq c 2^{2x} = c 2^{g(x)}$ whenever $x \geq -\log_2 c$. Therefore it is not true that $2^{f(x)} = \Omega(2^{g(x)})$ as $x \to \infty$.