Does $c^n = O(2^n)$ and $log_c(n) = O(log_2(n))$ for any constant $c$? I thought they did, but recently I tried to express $3^n$ as $k \times 2^n + o(2^n)$ for some constant $k$ but wasn't able to. All I found was $3^n = (\frac{3}{2})^n 2^n$. What am I misunderstanding here?
I suppose my question applies to $log_c(n) = O(log_2(n))$ as well.
Additionally to already answered in user5876164's comment let me say, that you cannot express $3^n=k \times 2^n+o(2^n)$, because if we assume it for some $k$, then we obtain: $\left(\frac{3}{2}\right)^n=k+o(1)$, which is impossible, as $\left(\frac{3}{2}\right)^n \to \infty$.