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Thread: Logarithm laws?

  1. #1
    Senior Member Twig's Avatar
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    Logarithm laws?

    Hi

    $\displaystyle ln(x^{2}) $ is defined for all $\displaystyle x \neq 0 $

    But, using basic logarithm law:

    $\displaystyle ln(x^{2}) = 2 \cdot ln(x) $

    ???
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Twig View Post
    Hi

    $\displaystyle ln(x^{2}) $ is defined for all $\displaystyle x \neq 0 $

    But, using basic logarithm law:

    $\displaystyle ln(x^{2}) = 2 \cdot ln(x) $

    ???
    ln(x) is still undefined for x=0 so I'm not sure what you're trying to ask
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  3. #3
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    Quote Originally Posted by Twig View Post
    $\displaystyle ln(x^{2}) $ is defined for all $\displaystyle x \neq 0 $
    But, using basic logarithm law:
    $\displaystyle ln(x^{2}) = 2 \cdot ln(x) $
    It works like this: $\displaystyle \sqrt {x^2 } = \left| x \right|$.
    Thus, $\displaystyle \ln \left( {x^2 } \right) = 2\ln \left( {\left| x \right|} \right)$.
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  4. #4
    Senior Member Twig's Avatar
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    Let $\displaystyle f(x) = ln(x^{2}) $

    Then, $\displaystyle D_{f}=\mathbb{R} $ , except $\displaystyle x = 0 $

    But, by writing $\displaystyle 2 \cdot ln(x) $ I dont have the same domain anymore. Now I need $\displaystyle x > 0 $
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