Why are different logarithms in the same Θ even thought their difference diverges?
You are missing one thing in your definition of asymptotic notation. In addition to only caring about large values of N, we also ignore constant multiplicative factors.

$f$ is $O(g)$ if there exists a large $N$ and a constant $c$ such that
$f(x) &lt; c g(x)$ for all $x \ge N$

In the $\log_3$ vs $\log_2$ case its true that we have $\log_3 x &lt; \log_2 x$ for all big $x$. However, its also true that

$\begin{equation*}
   \log_2 x
      = \frac{\log_3 x}{\log_2 3}
\end{equation*}$

so we can choose some $c$ larger than $\log_2 3$ to end up with:

$\begin{equation*}
    \log_2 x &lt; c \cdot \log_3 x
\end{equation*}$

and thus $\log_2 x$ is $O(\log_3 x)$.



The same reasoning does not apply to $2^n$ and $3^n$. There is no constant multiplicative factor that will make the $2^n$ catch up to $3^n$ for all big $n$.