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Math Help - Simply Fraction with negative exponets -Confused

  1. #1
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    Simply Fraction with negative exponets -Confused

    <br />
(\frac{2s^{-1}t^3}{6s^2t^{-4}})^{-3}<br />

    Which i reduced to ...
    <br />
\frac{27t^3}{s^3}<br />

    My work ...
    <br />
\frac{2^{-3}s^3t^{-9}}{6^{-3}s^{-6}t^{12}}<br />

    <br />
\frac{1}{8}*216=\frac{216}{8}=27<br />

    <br />
s^3*\frac{1}{s^6}=\frac{s^3}{s^6}=s^{-3}=\frac{1}{s^3}<br />

    <br />
\frac{1}{t^9}*t^{12}=\frac{t^{12}}{t^9}=t^3<br />

    Please help and thanks in advance.

    Paul
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by pkraus View Post
    <br />
(\frac{2s^{-1}t^3}{6s^2t^{-4}})^{-3}<br />

    Which i reduced to ...
    <br />
\frac{27t^3}{s^3}<br />

    My work ...
    <br />
\frac{2^{-3}s^3t^{-9}}{6^{-3}s^{-6}t^{12}}<br />

    <br />
\frac{1}{8}*216=\frac{216}{8}=27<br />

    <br />
s^3*\frac{1}{s^6}=\frac{s^3}{s^6}=s^{-3}=\frac{1}{s^3}<br />

    <br />
\frac{1}{t^9}*t^{12}=\frac{t^{12}}{t^9}=t^3<br />

    Please help and thanks in advance.

    Paul
    Personally I would solve it by flipping the fraction and making the power +3

    <br />
(\frac{2s^{-1}t^3}{6s^2t^{-4}})^{-3} = (\frac{6s^2t^{-4}}{2s^{-1}t^3})^3<br />

    = \frac{6^3s^6t^{-12}}{2^3s^{-3}t^9} = \frac{6^3}{2^3} \times \frac{3s^6}{s^{-3}} \times \frac{t^{-12}}{t^9} = 27s^9t^{-21}

    For non-zero a:
    a^{-n} = \frac{1}{a^n}

    a^na^m = a^{m+n}

    \frac{a^n}{a^m} = a^{n-m}

    In your case with s you've forgot to put the minus sign on the power in the denominator (see my red text)
    <br />
s^3*\frac{1}{s^{{\color{red}-}6}}=\frac{s^3}{s^{{\color{red}-}6}}=s^{3-{\color{red}(-6)}}=s^9<br />

    For t:

    From your original (and correct) expansion you got
    <br />
\frac{t^{-9}}{t^{12}}<br />

    However, in the next step you added the t^12 rather than subtracted it. Since t^12 is in the denominator it needs to have a negative power when multiplied (see the first law above)

    <br />
\frac{1}{t^9}*t^{{\color{red}-}12}=\frac{t^{{\color{red}-}12}}{t^9}=t^{-21}<br />
    Last edited by e^(i*pi); May 7th 2009 at 01:25 PM. Reason: Clarification
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by pkraus View Post
    <br />
(\frac{2s^{-1}t^3}{6s^2t^{-4}})^{-3}<br />

    Which i reduced to ...
    <br />
\frac{27t^3}{s^3}<br />

    My work ...
    <br />
\frac{2^{-3}s^3t^{-9}}{6^{-3}s^{-6}t^{12}}<br />

    <br />
\frac{1}{8}*216=\frac{216}{8}=27<br />

    <br />
s^3*\frac{1}{s^6}=\frac{s^3}{s^6}=s^{-3}=\frac{1}{s^3}<br />

    <br />
\frac{1}{t^9}*t^{12}=\frac{t^{12}}{t^9}=t^3<br />

    Please help and thanks in advance.

    Paul
    Hi Paul,

    Here's another take on it.

    Here's what I'd do with that. First invert the fraction so that the group will have a positive exponent.

    \left(\frac{2s^{-1}t^3}{6s^2t^{-4}}\right)^{-3}=\left(\frac{6s^2t^{-4}}{2s^{-1}t^3}\right)^3

    Next, simplify the group.

    \left(\frac{3s^3}{t^7}\right)^3

    Finally, bring it home.

    \frac{27s^9}{t^{21}}
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