Does $c^n = O(2^n)$ and $log_c(n) = O(log_2(n))$ for any constant $c$?
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 $\left(\frac{3}{2}\right)^n \to \infty$.