When are log complexities considered equivalent?
$O(\log(n))$ and $O(\log(n-1))$ contain the same set of functions.

The inclusion $O(\log(n-1)) \subseteq O(\log(n))$ is trivial.
As for $O(\log(n-1)) \supseteq O(\log(n))$, let $f(n) \in O(\log(n))$ and let $c \ge 1$ and $n_0 \ge 3$ be such that:
$$
f(n) \le c \log(n) \quad \forall n \ge n_0.
$$

We have:
$$
f(n) \le c \log(n) = c + c \log(n/2) \le c + c \log(n-1) \le 2c \log(n-1).
$$

Showing that $f(n) \in O(\log(n-1))$.