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Math Help - Another problem with fractional indices

  1. #1
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    Another problem with fractional indices

    I feel so stupid for having top ask for help again, but I have been trying to figure this out for several hours now




    2x^{\frac{1}{3}}={\frac{81}{8}}^{-1}

    I multiplied both sides by two to eliminate the fraction







    4x^{\frac{1}{3}}=27x^{-1}





    Then I multiplied both sides by x^{\frac{1}{3}}





    Result

    4=27^{\frac{-4}{3}}



    And I'm stuck.



    Would you say I have any hope of getting into Advanced Mathematics, if I can't even solve a problem like this?






    Thank you!
    Last edited by Coach; September 23rd 2007 at 06:16 AM.
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Coach View Post
    I feel so stupid for having top ask for help again, but I have been trying to figure this out for several hours now

    2x^{\frac{1}{3}}={\frac{81}{8}}^{-1}
    I looked at the raw Latex and it seems that the entire fraction on the right-hand side is supposed to be to the power of ^{-1}

    2x^{\frac{1}{3}}=( {\frac{81}{8}} ) ^{-1}

    2x^{\frac{1}{3}}= \frac{8}{81}

    x^{\frac{1}{3}}= \frac{4}{81}

    x= ( {\frac{4}{81}} ) ^\frac{3}{1}

    x = \frac{64}{531441}
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  3. #3
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    Thank you so much, but now I feel even more stupid, because I made a typo in the original question
    the correct one is supposed to be


     <br />
2x^{\frac{1}{3}}={\frac{81}{8}}x^{-1} <br />
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Coach View Post
    Thank you so much, but now I feel even more stupid, because I made a typo in the original question
    the correct one is supposed to be
    <br />
    " alt="2x^{\frac{1}{3}}={\frac{81}{8}}x^{-1}
    " />
     2x^{\frac{1}{3}}={\frac{81}{8}}x^{-1}

     x^{\frac{1}{3}}={\frac{81}{16}}x^{-1}

     ( x^{\frac{1}{3}} )( x ) = {\frac{81}{16}}

     ( x^{\frac{4}{3}} ) = {\frac{81}{16}}

     ( x ) = ( {\frac{81}{16}} )^{\frac{3}{4}}

     ( x ) = ( {\frac{27}{8}} )
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  5. #5
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    Thank you so much!
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Coach View Post
    Thank you so much!
    You're welcome and I look forward to helping you again
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