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Math Help - Extremely Simple Log Problem =D

  1. #1
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    Extremely Simple Log Problem =D

    How Would One Go about simplifying this ln(3x^2 - 9x) + ln(1/3x)
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  2. #2
    Super Member Bacterius's Avatar
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    Hi,
    you could use this law of logarithms : \log_a{(x)} + \log_a{(y)} = \log_a{(xy)}.

    But before proceeding to this, we need a little clarification : is your second term \ln{\left ( \frac{x}{3} \right )} or \ln{\left (\frac{1}{3x} \right )} ?
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  3. #3
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    <br /> <br />
\ln{\left (\frac{1}{3x} \right )}<br />
    It Looks like this
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  4. #4
    Super Member Bacterius's Avatar
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    Then, using the law of logarithms I mentioned in my previous post, you have :

    \ln{\left ( 3x^2 - 9x \right )} + \ln{\left ( \frac{1}{3x} \right )} = \ln{\left [ \left ( 3x^2 - 9x \right ) \left ( \frac{1}{3x} \right ) \right ]} = \ln{\left ( \frac{3x^2 - 9x}{3x} \right )}

    Note that the top part of the fraction, 3x^2 - 9x, can be factored as 3x(3x - 3). Therefore :

    \ln{\left ( \frac{3x^2 - 9x}{3x} \right )} = \ln{\left ( \frac{3x(3x - 3)}{3x} \right )} = \ln{\left ( 3x - 3 \right )}

    Conclusion : \ln{\left ( 3x^2 - 9x \right )} + \ln{\left ( \frac{1}{3x} \right )} = \ln{\left ( 3x - 3 \right )}

    Does this make sense ?
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  5. #5
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    Thank You very much very helpful
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