Is $\log(n-1) \in \Omega(\log(n))$?

Question

I saw this question Can I simplify log(n+1) before showing that it is in O(log n)? and wanted to know if a similar situation was also true.

Namely, is $\log(n-1) \in \Omega(\log(n))$?

Answer 1

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 $\alpha, \beta, \gamma \in \mathbb{R}_{>0}$.

Answer 2

Having $n-1 \gt \sqrt{n}$, when $n \gt 3$ we can write $$\log (n-1) \gt \log \sqrt{n} = \frac{1}{2}\log n$$ So, taking $N=3$ and $C=\frac{1}{2}\gt 0$ we fulfilled definition for $\Omega$.