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Thread: Expanding Log problem...

  1. #1
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    Expanding Log problem...

    Expand as the sum of individual logarithms, each of whose argument is linear: $\displaystyle log(\frac{xy^2}{z^4})$

    your help is appreciated! Thanks.
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  2. #2
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    Quote Originally Posted by Savior_Self View Post
    Expand as the sum of individual logarithms, each of whose argument is linear: $\displaystyle log(\frac{xy^2}{z^4})$

    your help is appreciated! Thanks.
    Using the following logarithm rules:

    $\displaystyle log(ab) = log(a) + log(b)$
    $\displaystyle log(\frac{a}{b})= log(a) - log(b)$
    $\displaystyle log(a^n) = n \cdot log(a)$

    I will start, see if you can finish:

    $\displaystyle log(\frac{xy^2}{z^4}) = log(xy^2) - log(z^4) = log(xy^2) - 4log(z)$

    ...
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    Using the following logarithm rules:

    $\displaystyle log(ab) = log(a) + log(b)$
    $\displaystyle log(\frac{a}{b})= log(a) - log(b)$
    $\displaystyle log(a^n) = n \cdot log(a)$

    I will start, see if you can finish:

    $\displaystyle log(\frac{xy^2}{z^4}) = log(xy^2) - log(z^4) = log(xy^2) - 4log(z)$


    ...
    so...

    $\displaystyle log(x) + log(y^2) - 4log(z)
    =
    log(x) + 2log(y) - 4log(z)$

    answer being...

    $\displaystyle log(x) + 2log(y) - 4log(z)$

    Look good?
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  4. #4
    Super Member
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    Israel
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    Quote Originally Posted by Savior_Self View Post
    so...

    $\displaystyle log(x) + log(y^2) - 4log(z)
    =
    log(x) + 2log(y) - 4log(z)$

    answer being...

    $\displaystyle log(x) + 2log(y) - 4log(z)$

    Look good?
    Yes, that is correct.
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