How to prove/disprove Big $\Theta$

Question

I would like to prove or disprove $$4^n = \Theta(2^n)$$

I think you may have to simplify the $4^n$ to $2^n*2^n$ but am unsure where to go from there.

Any idea? Thank you

Answer 1

Remember $ f(n) = \Theta(g(n)) $ if $ f(n) = O(g(n)) $ and $ f(n) = \Omega(g(n)) $

Also $ f(n) = O(g(n)) $ if $ f(n) < cg(n) $ for all $ n > n_0 $ and constant $ c $

With that, if $ 4^n = O(2^n) $, then $ 4^n < c2^n $ for all $ n > n_0 $ and constant $ c $, but this is clearly impossible because $ \frac{4^n}{2^n} < c $ and there cannot be such a constant.

So we have disproved $ 4^n = \Theta(2^n) $