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Math Help - Negative Exponents

  1. #1
    Mad
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    Negative Exponents

    (x^-1 y^2)^-3 (x^2 y^-4)^-3
    (x^3 y^-6)(x^-6 y^12)
    x^-3 y^6
    y^6/x^3


    Is that right?
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  2. #2
    Senior Member slevvio's Avatar
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    yup
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  3. #3
    Mad
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    (x^2 y^-1/xy)^-4
    (xy/x^2 y^-1)^4


    Why is it that when you flip fractions like that, nothing happens to the negative exponents?
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    Quote Originally Posted by Mad View Post
    (x^2 y^-1/xy)^-4
    (xy/x^2 y^-1)^4


    Why is it that when you flip fractions like that, nothing happens to the negative exponents?
    What do you mean that nothing happens to the negative exponents?
    x^{-a} = \frac{1}{x^a}

    In your case:
    \left ( \frac{x^2 y^{-1}}{xy} \right ) ^{-4}

    = \left ( \frac{xy}{x^2 y^{-1}} \right ) ^4

    I'm not sure why you are saying it hasn't changed?

    -Dan

    Edit:
    Oh wait! You are talking about the y^{-1} inside the expression, aren't you?
    \left ( \frac{\frac{x^2}{y}}{xy} \right ) ^{-4}

    = \left ( \frac{xy}{\frac{x^2}{y}} \right ) ^4

    The - in the -4 exponent flips the fraction in the expression over. The numerator of the expression contains a fraction, so this doesn't change.

    This is why I think it is better to apply the multiplication rule of exponents first, then simplify.

    -Dan
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  5. #5
    Mad
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    Thanks.


    One more question.

    (x^2y^-1/xy)^-4
    (xy/x^2y^-1)^4

    My professor would have simplified what's inside the parantheses first. She would have done 1 - 2 for the x and got -1. She would have taken 1 - (-1) for y and gotten 2. I think.

    I think this is what I used to do in high school:

    x : 2 - 1 = 1; the x on the bottom is bigger so that's where the x remains
    y : 1 - 1 = 0; no y

    Right or wrong?


    I kind of think I'm wrong but I'm confused...
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  6. #6
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    Quote Originally Posted by Mad View Post
    Thanks.


    One more question.

    (x^2y^-1/xy)^-4
    (xy/x^2y^-1)^4

    My professor would have simplified what's inside the parantheses first. She would have done 1 - 2 for the x and got -1. She would have taken 1 - (-1) for y and gotten 2. I think.

    I think this is what I used to do in high school:

    x : 2 - 1 = 1; the x on the bottom is bigger so that's where the x remains
    y : 1 - 1 = 0; no y

    Right or wrong?


    I kind of think I'm wrong but I'm confused...
    Compare the marked calcualtions!
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  7. #7
    Mad
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    Yeah, I noticed that. I think I know where I went wrong -- I remembered my high school teacher talking about subtracting the two and putting the answer where the biggest number had been before but forgot about signs and such.

    So you subtract the bottom number from the top? And then what? The answer goes on the top? But if it's a negative, you put it on the bottom?

    If so, I'll just have to throw out that whole "Put the answer where the biggest number was before." thing.


    EDIT: While I'm at it...

    y^8/x^4? Is that right?
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  8. #8
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    Quote Originally Posted by Mad View Post
    ...

    So you subtract the bottom number from the top? And then what? The answer goes on the top? But if it's a negative, you put it on the bottom?

    If so, I'll just have to throw out that whole "Put the answer where the biggest number was before." thing.

    EDIT: While I'm at it...
    y^8/x^4? Is that right?
    I assume that you know the basic properties of powers:

    a^n \cdot a^m = a^{n+m}

    a^{-n} = \frac1{a^n}

    \left(a^n\right)^m = a^{n \cdot m}

    Apply these rules to your question and transform the complete term into a product:

    \left(\frac{x^2 \cdot y^{-1}}{x \cdot y} \right)^{-4} = \left(x^2 \cdot y^{-1} \cdot \frac1x \cdot \frac1y \right)^{-4} = \left(x^2 \cdot \frac1x \cdot y^{-1}   \cdot \frac1y \right)^{-4} = \left(x^2 \cdot x^{-1} \cdot y^{-1}   \cdot y^{-1} \right)^{-4} = \left(x^{2-1} \cdot y^{-1-1} \right)^{-4} = (x^1 \cdot y^{-2})^{-4} = x^{-4} \cdot y^8 = \frac{y^8}{x^4}

    So your answer is correct
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