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Math Help - Some help on logarithmic laws

  1. #1
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    Some help on logarithmic laws

    Hello,

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

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


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

    Thank you in advance!
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Some help on logarithmic laws

    Write:
    \ln \left(1+\frac{1}{n}\right)=-\ln \left(1-\frac{1}{n+1}\right) as
    \ln \left(\frac{n+1}{n}\right)=-\ln \left(\frac{n}{n+1}\right)
    And we know (in general): y\log_a(x)=\log_a(x^y), so ...
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  3. #3
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    Re: Some help on logarithmic laws

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

    1+ \frac{1}{n}= \frac{n}{n}+ \frac{1}{n}= \frac{n+1}{n}
    while
    1- \frac{1}{n+1}= \frac{n+1}{n+1}- \frac{1}{n+1}= \frac{n}{n+1}
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  4. #4
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    Re: Some help on logarithmic laws

    Quote Originally Posted by Melsi View Post
    \ but cannot find which law justifies this:
    ln(1+\frac{1}{n}) = -ln(1-\frac{1}{n+1})
    Just take note that \left( {1 + \frac{1}{n}} \right)^{ - 1}  = \left( {\frac{n}{{n + 1}}} \right) = \left( {1 - \frac{1}{{n + 1}}} \right)
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  5. #5
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    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!
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