# Thread: Some help on logarithmic laws

1. ## Some help on logarithmic laws

Hello,

I am looking up different logarithmic tutorials but cannot find which law justifies this:

$\displaystyle ln(1+\frac{1}{n}) = -ln(1-\frac{1}{n+1})$

could $\displaystyle ln x \leq (x-1)$ be of any help?

2. ## Re: Some help on logarithmic laws

Write:
$\displaystyle \ln \left(1+\frac{1}{n}\right)=-\ln \left(1-\frac{1}{n+1}\right)$ as
$\displaystyle \ln \left(\frac{n+1}{n}\right)=-\ln \left(\frac{n}{n+1}\right)$
And we know (in general): $\displaystyle y\log_a(x)=\log_a(x^y)$, so ...

3. ## Re: Some help on logarithmic laws

log(1/x)= -log(x) together with basic algebra.

$\displaystyle 1+ \frac{1}{n}= \frac{n}{n}+ \frac{1}{n}= \frac{n+1}{n}$
while
$\displaystyle 1- \frac{1}{n+1}= \frac{n+1}{n+1}- \frac{1}{n+1}= \frac{n}{n+1}$

4. ## Re: Some help on logarithmic laws

Originally Posted by Melsi
\ but cannot find which law justifies this:
$\displaystyle ln(1+\frac{1}{n}) = -ln(1-\frac{1}{n+1})$
Just take note that $\displaystyle \left( {1 + \frac{1}{n}} \right)^{ - 1} = \left( {\frac{n}{{n + 1}}} \right) = \left( {1 - \frac{1}{{n + 1}}} \right)$

5. ## Re: Some help on logarithmic laws

Thank you all guys, you've been so helpful. I am glad that appart from algeraic solution a hint to the specific logarithmic law was mentioned!

Bye every one!