Is $\log(n-1) \in \Omega(\log(n))$?
As you have seen, $\log(n+1) \in \mathcal{O}(\log n)$. That means that $\log n \in \Omega(\log( n + 1))$. Now you can change $n$ with $n-1$ and have your answer.
To be more precise, $\log(\alpha n^{\beta} + \gamma) \in \Theta(\log n)$ for any $\alpha, \beta, \gamma \in \mathbb{R}_{&gt;0}$.