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Exponent laws

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  1. Jan 28, 2016 #1
    1. The problem statement, all variables and given/known data
    x(cnxn-1)

    3. The attempt at a solution

    I know that the answer is cnxn

    I'm not sure why though. My thinking is that we have cnxn-1 and we multiply that by x. x times x is x2 so I'm expecting a 2 to interact with the n-1 in the exponents. I'm just not sure how n-1 interacts with 2 to produce n.

    I don't add n-1 + 2 to get n + 1 because I already have the answer and its not that. I just can't see how I would get from n-1 to n by interacting with a 2.

    Thanks for the help.
     
  2. jcsd
  3. Jan 28, 2016 #2

    SammyS

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    You may know the answer to the question.

    We're not any good at guessing just what that question might be.

    What is the question?
     
  4. Jan 28, 2016 #3
    The question is the multiplication of x times cnxn-1
     
  5. Jan 28, 2016 #4

    SteamKing

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    I don't know where you are getting that you are multiplying x times x here.

    You are multiplying x times xn-1, which is something different.

    That's what x(cnxn-1) means. It can also be written as c ⋅ n ⋅ x ⋅ xn-1
     
  6. Jan 28, 2016 #5

    SammyS

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    Oh! You want to simplify the product, x⋅cnxn-1 .

    So the 'question' is:
    What is the result of simplifying x⋅cnxn-1 ?
    And you want to know how it is that simplifies to be cnxn .
    .
     
  7. Jan 28, 2016 #6

    HallsofIvy

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    Have you never actually taken an Algebra class? If you have you should have learned one of the basic 'laws of exponents':
    [itex](a^m)(a^n)= a^{m+n}[/itex]. Here the problem is to multiply x, which is the same as [itex]x^1[/itex], by [itex]x^{n-1}[/itex].
    Mod note: Portion deleted as too much help.
     
    Last edited by a moderator: Jan 28, 2016
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