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Math Help - How do you solve this logarithm?

  1. #1
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    How do you solve this logarithm?

    If ln a = 2, ln b = 3, and ln c = 5, evaluate:

    1. ln((a^-4)(b^1)) / ln(bc)^3

    and

    2. (ln c^-1)(ln a/(b^3))^4
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  2. #2
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    Recall and apply your log rules.

    log (AB) = log A + log B

    log(A/B) = log A - log B

    log (A)^n = n logA
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  3. #3
    Member rtblue's Avatar
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    In the following lines, I mean to say ln(a^-4), not ln(a-4). Latex issues, sorry.

    \frac{ln[(a^-4)(b)]}{ln(bc)^3}

    \frac{ln(a^-4)+ln(b)}{3ln(bc)}

    \frac{-4ln(a)+ln(b)}{3ln(b)+3ln(c)}

    \frac{-4(2)+3}{3(3)+3(5)}

    \frac{-5}{24}
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by rtblue View Post
    In the following lines, I mean to say ln(a^-4), not ln(a-4). Latex issues, sorry.

    \frac{ln[(a^-4)(b)]}{ln(bc)^3}

    \frac{ln(a^-4)+ln(b)}{3ln(bc)}

    \frac{-4ln(a)+ln(b)}{3ln(b)+3ln(c)}

    \frac{-4(2)+3}{3(3)+3(5)}

    \frac{-5}{24}
    FYI you can use [tex]\ln \left(a^{-4}\right)[/tex] which gives \ln \left(a^{-4}\right)

    I've nothing new to contribute to the thread, look at the two posts above me
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