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Thread: confused... fractions and indicies

  1. #1
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    confused... fractions and indicies

    Simplify


    $\displaystyle 16^1/2/81^3/4$


    i got the answer again but im not sure about the bottom half since i just cube rooted the 81 and wasnt sure what to do with the 4 part of it.... dont know if this is coincidently the right answer that i got...

    please explain the method, thanks!

    sorry i cant get the hang of the [tex] thing , its suppose to be 16^half all over (divided by) 81^three quarters
    Last edited by coyoteflare; Jan 8th 2009 at 09:37 AM. Reason: apologies
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  2. #2
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by coyoteflare View Post
    Simplify


    $\displaystyle 16^1/2/81^3/4$


    i got the answer again but im not sure about the bottom half since i just cube rooted the 81 and wasnt sure what to do with the 4 part of it.... dont know if this is coincidently the right answer that i got...

    please explain the method, thanks!
    I don't get it... you mean...

    $\displaystyle 16$ divided by $\displaystyle 2$ divided by $\displaystyle 18^{3}$ divided by $\displaystyle 4$?

    Edit: Ow like this?

    $\displaystyle \sqrt{16} = 16^{\frac{1}{2}}$

    $\displaystyle 16^{\frac{1}{2}} = 4$

    $\displaystyle \sqrt[4]{81^{3}} = 81^{\frac{3}{4}}$

    $\displaystyle 81^{\frac{3}{4}} = 27$

    $\displaystyle \frac{16^{\frac{1}{2}}}{81^{\frac{3}{4}}}= \frac{4}{27}$
    Last edited by shinhidora; Jan 8th 2009 at 09:57 AM.
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  3. #3
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    $\displaystyle \frac{16^{1/2}}{81^{3/4}}$



    16 to the power of a half divided by 81 to the power of three quarters

    i cant get the Latex to do it tooconfusing for me
    Last edited by mr fantastic; Jan 8th 2009 at 12:01 PM. Reason: Fixed the latex
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  4. #4
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by coyoteflare View Post
    [tex]{\frac{16^{\frac{1}{2}}}{81^{\frac{3}{4}}}



    16 to the power of a half divided by 81 to the power of three quarters

    i cant get the Latex to do it tooconfusing for me

    It isn't that hard

    In short, for your problem

    \frac{a}{b} = $\displaystyle \frac{a}{b}$
    a^{b} = $\displaystyle a^{b}$

    By the way, I edited my original post with the answers and stuff
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  5. #5
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    Quote Originally Posted by shinhidora View Post
    I don't get it... you mean...

    $\displaystyle 16$ divided by $\displaystyle 2$ divided by $\displaystyle 18^{3}$ divided by $\displaystyle 4$?

    Edit: Ow like this?

    $\displaystyle \frac{16^{\frac{1}{2}}}{81^{\frac{3}{4}}}$
    That's what I understood, too. Using radicals it would be

    $\displaystyle \frac{\sqrt{16}}{(\sqrt[4]{81})^3}$
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  6. #6
    Junior Member shinhidora's Avatar
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    In general it's

    $\displaystyle \sqrt[a]{n^{b}} = n^{\frac{b}{a}}$


    This will make it more clear I guess
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  7. #7
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    without soundign too stupid

    how is$\displaystyle

    \sqrt[4]{81^{3}} = 81^{\frac{3}{4}}
    $ = 27?

    i do $\displaystyle \sqrt[4]{81}$ then cube the answer and it coems out massive!?
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  8. #8
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    Quote Originally Posted by coyoteflare View Post
    how is$\displaystyle

    \sqrt[4]{81^{3}} = 81^{\frac{3}{4}}
    $ = 27?

    i do $\displaystyle \sqrt[4]{81}$ then cube the answer and it coems out massive!?
    Hello coyote,

    The 4th root of 81 is 3.

    3 cubed = 27. Not too massive.
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