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Evaluating rational exponents

  1. Oct 4, 2006 #1
    Can anyone point me to a text or link that summarizes the rules when evaluating/simplifying an expression of the form
    [tex] (a^n)^(1/m) [/tex] for a < 0. [tex] (a^n)^(1/m) [/tex] yields different answers for [tex] a^(n/m) [/tex] and [tex] (a^(1/m))^n [/tex].

    Ex:

    [tex] (-8)^(2/6) = (-8)^(1/3) = -2 [/tex]
    [tex] (-8)^(2/6) = ((-8)^2)^(1/6) = 2 [/tex]
    [tex] (-8)^(2/6) = (-8^(1/6))^2 = undefined [/tex]

    Thank you very much!
     
    Last edited: Oct 4, 2006
  2. jcsd
  3. Oct 4, 2006 #2

    HallsofIvy

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    The general function, ax, is only defined for positive a.
    There simply are no ways of giving "rules" for such manipulation with a negative.
     
  4. Oct 4, 2006 #3
    Thanks HallsofIvy

    i've read in some book that
    [tex] (a^n)^{1/m} [/tex]
    where a < 0, n and m are positive even integers and [tex] a^{1/m} [/tex] is defined can be simplified to
    [tex] |a|^{n/m} [/tex]

    There are just a few examples on this subject that's why i'm looking for other resources. I've solved several exercises and I different answers. I miss out on when to place the absolute value bars and when not to. As I understand it, the absolute value bars may be removed if n/m always yields a positive value for [tex] |a|^{n/m} [/tex], otherwise, the absolute value bars must be retained.

    I know this is elementary for you guys. :biggrin:
     
    Last edited: Oct 5, 2006
  5. Oct 5, 2006 #4

    Integral

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    Just a LaTex note. surround your exponent with curly brackets {} to get it all elevated.\

    [tex] a ^{ \frac 1 m} [/tex]

    click on the equation to see the code.
     
  6. Oct 5, 2006 #5
    Just a rewrite

    Can anyone point me to a text or link that summarizes the rules when evaluating/simplifying an expression of the form
    [tex] (a^n)^{1/m} [/tex] for a < 0. [tex] (a^n)^{1/m} [/tex] yields different answers for [tex] a^{n/m} [/tex] and [tex] (a^{1/m})^n [/tex].

    Ex:

    [tex] (-8)^{2/6} = (-8)^{1/3} = -2 [/tex]
    [tex] (-8)^{2/6} = ((-8)^2)^{1/6} = 2 [/tex]
    [tex] (-8)^{2/6} = ((-8)^{1/6})^2 = undefined [/tex]

    Thank you very much!
     
    Last edited: Oct 5, 2006
  7. Oct 5, 2006 #6

    VietDao29

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    No, that's not true. There's no such way to simplify that. :frown: :smile:
     
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