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How do I solve (a+b)^(-c)?

  1. Sep 24, 2010 #1
    Good day,

    How do I work out (a+b)^(-c)?

    Thanks.
     
  2. jcsd
  3. Sep 24, 2010 #2
    Re: (a+b)^(-c)

    In my case, c = 1, so I have got: (a+b)^-1
     
  4. Sep 24, 2010 #3

    statdad

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    Homework Helper

    Re: (a+b)^(-c)

    Well, what does [itex] x ^{-1} [/itex] represent?
     
  5. Sep 24, 2010 #4
    Re: (a+b)^(-c)

    (1/x).

    I have got the equation:
    r = 1/(a + bcos(c))

    This should be equal to:
    r = (1/a) * (1/(1+bcos(c)))

    I just can't figure out why.
     
  6. Sep 24, 2010 #5

    statdad

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    Homework Helper

    Re: (a+b)^(-c)

    To be straight that I understand: you have

    [tex]
    r = \frac 1 {a + b \cos{c}}
    [/tex]

    and need to show that this equals

    [tex]
    r = \left(\frac 1 a\right) \left( \frac 1 {1 + b \cos{c}}\right)
    [/tex]

    If your statements are the ones I've written here, they aren't equal.
     
  7. Sep 24, 2010 #6
    Re: (a+b)^(-c)

    Exactly.
     
  8. Sep 24, 2010 #7
    Re: (a+b)^(-c)

    Are there some type of conditions? If not, just write down a counterexample, therefore showing it's impossible to prove it (ie. a not = 1, b, c in reals)
     
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