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Math Help - A problem with fractional indices

  1. #1
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    A problem with fractional indices

    I have tried everything I know, but I can't solve this. Can you please show me the right direction?


    \frac{16^\frac{1}{3}*4^\frac{1}{3}}{8}













    Thank you!
    Last edited by Coach; September 9th 2007 at 11:49 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Coach View Post
    I have tried everything I know, but I can't solve this. Can you please show me the right direction?


    \frac{16^\frac{1}{3}*4^\frac{1}{3}}{8}
    change everything to base 2

    \frac {16^{\frac {1}{3}} \cdot 4^{\frac {1}{3}}}{8} = \frac { \left( 2^{4} \right)^{\frac {1}{3}} \cdot \left( 2^{2} \right)^{\frac {1}{3}}}{2^3}

    Now continue
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  3. #3
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    Thank you so much!


    But I would still have another question

    \frac{x^{2n+1}*x^\frac{1}{2}}{\sqrt{x^{3n}}}










    Thank you!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Coach View Post
    Thank you so much!


    But I would still have another question

    \frac{x^{2n+1}*x^\frac{1}{2}}{\sqrt{x^{3n}}}
    change the square root in the denominator to a power (you know how to do that, right? see my post here)

    so we get \frac{x^{2n+1}*x^\frac{1}{2}}{\sqrt{x^{3n}}} = \frac{x^{2n+1}*x^\frac{1}{2}}{x^\frac {3n}{2}}

    Now continue
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  5. #5
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    ok, so is it like

    x^{2n+1+\frac{1}{2}}-^{\frac{3n}{2}} ?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Coach View Post
    ok, so is it like

    x^{2n+1+\frac{1}{2}}-^{\frac{3n}{2}} ?
    yes
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