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

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

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 $$\Omega$$.
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 $$\alpha, \beta, \gamma \in \mathbb{R}_{>0}$$.
• Hope Your $\mathbb{R}_+^*$ doesn't contain $\alpha=0$, as then we have constant on left side. Feb 24 at 10:47