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Math Help - Logarithms

  1. #1
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    Johor Bahru
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    Logarithms

    Simplify, 3 + 2log(base p+q)(1/p^2-q^2)
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  2. #2
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    You can factorise
    1/(P^2 - Q^2) to 1/(P + Q)(P - Q)

    Then you get 3 + log(1/(P-Q). You can make this 3+ log1 - log(P-Q) = 3 + 0 - log(P-Q)

    Oh: And I refuse 2 use that math text app.. I don't understand it, some crazy foreign stuff imo!
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  3. #3
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    Quote Originally Posted by nav92 View Post
    Simplify, 3 + 2log(base p+q)(1/p^2-q^2)
    3 - 2\log_{p+q}\left[(p+q)(p-q)\right]

    3 - 2\log_{p+q}(p+q) - 2\log_{p+q}(p-q)

    3 - 2 - 2\log_{p+q}(p-q)

    1 - \log(p-q)^2
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  4. #4
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    hmm, I think he meant log(a*b) where a was P+Q and b was 1/(P^2 - Q^2)
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  5. #5
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    Quote Originally Posted by narko View Post
    hmm, I think he meant log(a*b) where a was P+Q and b was 1/(P^2 - Q^2)
    note the word "base" by (p+q)
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