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Double Exponents Question

  1. Mar 16, 2013 #1
    Hi,
    If all x,a,b and c are all natural numbers, is this true?
    [itex] x^{a^b} = (x^{a^{b-1}})^a[/itex]
    Proof
    if [itex] c = a^{b-1}[/itex]
    [itex] ca = (a^{b-1})a = a^b [/itex]
    and [itex] (x^c)^a = x^{ca} = x^{a^b} [/itex]

    Could I please have some feedback on this,
    Thanks
     
  2. jcsd
  3. Mar 16, 2013 #2

    pwsnafu

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    Science Advisor

    There is no universal agreement on whether zero is a natural number. You are going to need write "positive integer" or deal with the case when one or a number of the variables are zero separately.
     
    Last edited: Mar 16, 2013
  4. Mar 17, 2013 #3

    Mentallic

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    Homework Helper

    Yes, and the proof doesn't require any substitutions either. Simply following your exponent laws,

    [tex]\left(x^{a^{b-1}}\right)^a[/tex]
    [tex]=x^{a^{b-1}a}[/tex]
    [tex]=x^{a^{b-1+1}}[/tex]
    [tex]=x^{a^b}[/tex]
     
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