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Math Help - How am I supposed to know that ln9 - ln1 = 2*ln3 ?

  1. #1
    s3a
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    How am I supposed to know that ln9 - ln1 = 2*ln3 ?

    How am I supposed to know that ln9 - ln1 = 2*ln3 ? For example, I would know that 9/4 - 3/2 is 15/4 because I can convert 3/2 into 6/4 but how do I such a thing with logarithms?

    Any input would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by s3a View Post
    How am I supposed to know that ln9 - ln1 = 2*ln3 ? For example, I would know that 9/4 - 3/2 is 15/4 because I can convert 3/2 into 6/4 but how do I such a thing with logarithms?

    Any input would be greatly appreciated!
    Thanks in advance!
    From the basics of logs you're expected to know that \log_b(1) = 0 \: \: \: , \: \: b > 1. This is because b^0 = 1

    This expression is simplified using the log power law: \ln(a^k) = k\, ln(a) \: \: , \: \: a > 0 and the knowledge of the fact that 9=3^2

    Thus \ln(9) - \ln(1) = \ln(3^2) - 0 = 2\ln(3)


    ====================

    Laws of logarithms. b > 1 in all cases and a,c > 0 in all cases

    \log_b(a) + \log_b(c) = \log_b(ac)

    \log_b(a) - log_b(c) = \log_b \left(\frac{a}{c}\right)

    \log_b(b) = 1

    \log_b(a^k) = k\,\log_b(a)

    \log_b \left(\frac{1}{a^k}\right) = -k\,\log_b(a)

    \log_b(1) = 0

    \log_b(a) = c  \: \: \rightarrow \: \: a = b^c
    Last edited by e^(i*pi); January 30th 2010 at 08:07 AM. Reason: adding latex/log laws
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