# Logarithms and factorial; identity

• Jun 29th 2011, 04:44 AM
Zellator
Logarithms and factorial; identity
Hi Forum
Can someone help?
There's a question that states

$\displaystyle 1+\log_{n!} {(n+1)} = \log_{n!} {(n+1)!}$

The question is actually
$\displaystyle \displaystyle\prod_{n=2}^{10} (1+\log_{n!} {(n+1)})$

Does it change anything by being Cartesian product?
Thanks:)
• Jun 29th 2011, 04:58 AM
Soroban
Re: Logarithms and factorial; identity
Hello, Zellator!

Quote:

$\displaystyle 1+\log_{n!}(n+1) \:=\: \log_{n!}(n+1)!$

$\displaystyle \text{But how was this thought?}$

Note that: .$\displaystyle \log_{n!}(n!) \:=\:1$

The left side becomes:
. . $\displaystyle \log_{n!}(n!) + \log_{n!}(n+1) \;=\;\log_{n!}\bigg[n!(n+1)\bigg] \;=\;\log_{n!}(n+1)!$

• Jun 29th 2011, 05:15 AM
Re: Logarithms and factorial; identity
Quote:

Originally Posted by Zellator
Hi Forum
Can someone help?
There's a question that states

$\displaystyle 1+\log_{n!} {(n+1)} = \log_{n!} {(n+1)!}$

The question is actually
$\displaystyle \displaystyle\prod_{n=2}^{10} (1+\log_{n!} {(n+1)})$

Does it change anything by being Cartesian product?
Thanks:)

Working from the right

$\displaystyle log_{n!}(n+1)!=log_{n!}\left[(n+1)n!\right]=log_{n!}(n!)+log_{n!}(n+1)$
• Jun 29th 2011, 05:37 AM
Zellator
Re: Logarithms and factorial; identity
Quote:

Originally Posted by Soroban
Hello, Zellator!

Note that: .$\displaystyle \log_{n!}(n!) \:=\:1$

The left side becomes:
. . $\displaystyle \log_{n!}(n!) + \log_{n!}(n+1) \;=\;\log_{n!}\bigg[n!(n+1)\bigg] \;=\;\log_{n!}(n+1)!$

Then using $\displaystyle n!(n+1)$ we can see that $\displaystyle (n+1)(n)(n-1)!$
So $\displaystyle n!(n+1)= (n+1)!$

But that's kind of hard to see in the naked eye, thanks to you I will probably find it this forms easily from now on.

Quote:

Working from the right

$\displaystyle log_{n!}(n+1)!=log_{n!}\left[(n+1)n!\right]=log_{n!}(n!)+log_{n!}(n+1)$

The thing is, that the right form wasn't given, it had to be calculated, in order to find a better and easier way to solve the Cartesian Product.
:D
• Jun 29th 2011, 07:12 AM
Re: Logarithms and factorial; identity
Ok,

well that's important to realise about factorials.

$\displaystyle 6!=(6)5!$

$\displaystyle 7!=(7)6!$

and so on.

Also, a very fundamental property of logarithms is

$\displaystyle log_{a}a=y\Rightarrow\ a=a^y\Rightarrow\ y=1$

Then

$\displaystyle 1+log_{n!}(n+1)=1+log_{n!}\frac{n!(n+1)}{n!}=1+log _{n!}\frac{(n+1)!}{n!}$

$\displaystyle =1+log_{n!}(n+1)!-log_{n!}(n!)$