How can I prove that this is true/false:
$$\log(3^n) \in O(\log(2^n))\ ?$$
I know $f(n)$ is $O(g(n))$ if there are positive constants $C$ and $k$ such that: $$f(n) \le C \cdot g(n)$$ whenever $n > k$.
I think that the plot shows that it is not in $O(\log(2^n))$. Can I prove it with $$0 \leq \lim\limits_{n \to \infty} \dfrac{g(n)}{f(n)} < \infty\ ?$$ Thanks in advance!
EDIT: Thanks, I found the answer, the hint that it is a linear function help me! It is true for big-O and also true for big-$\Omega$.
$$ \log(3^n) = n \log(3) = n \log(3) \frac{\log(2)}{\log(2)} = n \log(2) \frac{\log(3)}{\log(2)} = C \log(2^n) $$