why $\log(n!)$ isn't zero?

Question

I have wondered that why the $\log (n!)$ isn't zero for $n \in N$.

Because I think that $\log (1)$ is zero so all following numbers after multiplying the result will become zero.

Thanks in advance.

to format mathematics insert the text code between dollar signs: If you meant to write $\log(n)!$ then you type \log(n)! between dollar signs. Try it! — Ittay Weiss, Mar 31 '13 at 05:25
Do you mean $\lg(n!)$? That’s $\lg 1+\lg 2+\ldots+\lg n$: you’re adding the logs, not multiplying them. — Brian M. Scott, Mar 31 '13 at 05:25
Oh, sorry for the stupid question. I got it and you can answer it so that I can accept your answer. Thank you! — Jason, Mar 31 '13 at 05:28
I think you mean $\log (n!)$ with the exclamation mark inside the parentheses, which was Brian's suggestion. — Will Jagy, Mar 31 '13 at 05:29

score 18 · Accepted Answer · answered Mar 31 '13 at 05:30

18

Might as well make an answer of it.

$$\begin{align*} \lg(n!)&=\lg(1\cdot2\cdot3\cdot\ldots\cdot n)\\ &=\lg 1+\lg 2+\lg 3+\ldots+\lg n\\ &=\lg 2+\lg 3+\ldots+\lg n\;, \end{align*}$$

so it won’t be $0$ unless $n=1$ (or $n=0$): you’re adding the logs, not multiplying them.

answered Mar 31 '13 at 05:30

Brian M. Scott

1 Answers1