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Math Help - Logarithm laws?

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

    Hi

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

    But, using basic logarithm law:

     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

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

    But, using basic logarithm law:

     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
     ln(x^{2}) is defined for all  x \neq 0
    But, using basic logarithm law:
     ln(x^{2}) = 2 \cdot ln(x)
    It works like this: \sqrt {x^2 }  = \left| x \right|.
    Thus, \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  f(x) = ln(x^{2})

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

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