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Math Help - Laws of exponents

  1. #1
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    Laws of exponents

    this is the question i am having problems with:

    [(x^3y^-2) / (x^-1y^4)]^-1/2 [x^-1/2y^3/4]^-1

    Sorry if this is unclear but I don't know how to represent it in any other way haha
    this is what i did:

    [ (x^-3/2y^1) /(x^1/2y^-2)] [x^1/2y^-3/4]
    = [x^-2y^3][x^1/2y^-3/4]
    = x^-3/2y^9/4

    The answer in the back is x^-3/2y^15/4
    So, I got the x part right but not the y, i don't see how you would be able to add 3 and -3/4 as one of them is negative so you would evenutally have to subtract.

    is the answer in the back just wrong?
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  2. #2
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    It looks like in your second bracket, that y^(3/4) (You really must use parentheses around exponents!) is in the denominator. If so, then when you put it in the numerator, you have y^(-3/4), and then raised to the (-1) power gets you y^(+3/4). So, you're going to take y^(3)y^(+3/4) = y^(15/4).

    Make sense?
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  3. #3
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    Thanks, yes i do think i understand. not sure why you don't do that to other exponents in the question though like the x^-2 and such..
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  4. #4
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    Quote Originally Posted by tmas View Post
    not sure why you don't do that to other exponents in the question though like the x^-2 and such..
    To what are you referring? You have this:

    \left(\frac{x^{3}y^{-2}}{x^{-1}y^{4}}\right)^{-1/2}\left(\frac{x^{-1/2}}{y^{3/4}}\right)^{-1}=(x^{3}y^{-2}x^{1}y^{-4})^{-1/2}(x^{-1/2}y^{-3/4})^{-1}=(x^{4}y^{-6})^{-1/2}(x^{-1/2}y^{-3/4})^{-1}

    =x^{-2}y^{3}x^{1/2}y^{3/4}...
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  5. #5
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    \frac{}{ }
    Quote Originally Posted by Ackbeet View Post
    To what are you referring? You have this:

    \left(\frac{x^{3}y^{-2}}{x^{-1}y^{4}}\right)^{-1/2}\left(\frac{x^{-1/2}}{y^{3/4}}\right)^{-1}=(x^{3}y^{-2}x^{1}y^{-4})^{-1/2}(x^{-1/2}y^{-3/4})^{-1}=(x^{4}y^{-6})^{-1/2}(x^{-1/2}y^{-3/4})^{-1}

    =x^{-2}y^{3}x^{1/2}y^{3/4}...
    the question that you started with is not similar to the one i started with, in the second bracket the y^3/4 is not in the demonimator but the whole bracket itself is the numerator (aka everything is over 1)

    I tried to do this a different way where while dealing with the exponents in the 2nd bracket i put everything over 1 (since the exponent is -1) therefore making that second bracket [1/ (x^1/2y^-3/4)] i changed the 1/2 power from negative to positive and the 3/4 power from postive to negative after that i just multiplied across with exponents like this:

    [ (x^-3/2y^1) / (x^1/2y^-2) ] TIMES [ 1 / (x^1/2y^-3/4)]

    = (x^-3/2y) / (xy^-11/4)

    = x^-5/2y^15/4

    SO NOW, when i do it this way i get the correct exponent values for the y and incorrect for the x.. but in the step before the x is the correct value.

    Ugh. I really do not understand what I am doing wrong and am about to give up on this question.
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  6. #6
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    Quote Originally Posted by tmas View Post
    \frac{}{ }

    the question that you started with is not similar to the one i started with, in the second bracket the y^3/4 is not in the demonimator but the whole bracket itself is the numerator (aka everything is over 1)
    I assume you actually meant to say that "the whole bracket itself is the denominator (aka, 1 is over everything)." In that case, what I wrote is the same thing as what you just said, since

    \frac{1}{x}=x^{-1}.

    In this context, that means

    \left(\frac{x^{-1/2}}{2y^{3/4}}\right)^{-1}=\frac{1}{\frac{x^{-1/2}}{2y^{3/4}}}.

    It is NOT true that

    \left(\frac{x^{-1/2}}{2y^{3/4}}\right)^{-1}=\frac{1}{\frac{x^{1/2}}{2y^{-3/4}}},

    for so many reasons. For one thing, you've kind of flipped the exponent signs twice, where you were only allowed to do it once. For another, you've distributed the exponents to everything except the 2.

    I tried to do this a different way where while dealing with the exponents in the 2nd bracket i put everything over 1 (since the exponent is -1) therefore making that second bracket [1/ (x^1/2y^-3/4)] i changed the 1/2 power from negative to positive and the 3/4 power from postive to negative after that i just multiplied across with exponents like this:

    [ (x^-3/2y^1) / (x^1/2y^-2) ] TIMES [ 1 / (x^1/2y^-3/4)]

    = (x^-3/2y) / (xy^-11/4)

    = x^-5/2y^15/4

    SO NOW, when i do it this way i get the correct exponent values for the y and incorrect for the x.. but in the step before the x is the correct value.

    Ugh. I really do not understand what I am doing wrong and am about to give up on this question.
    I think you're making more work for yourself, and therefore more opportunities to make a mistake. Take it one step at a time. If I were you, I'd get everything in the numerator first, like I did. Flip exponent signs as needed to do so. Then distribute the outside exponents, and so on.
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  7. #7
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    no actually i meant what i said.

    the original question for the second bracket states (x^-1/2y^3/4)^-1

    The only thing that is tripping me up about your explanation is when you show me that portion as (x^-1/2) / (y^3/4) so when you do the reciprocal of that i would get what you got but the question is not presented that way.

    Also, i'm confused where the two came from.

    I'll just ask my teacher next class. cause there is clearly a void in the communication here..

    Thanks for your help.
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  8. #8
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    Here's my confusion - I think I finally figured it out. It's from your lack of parentheses. In Post # 2, I warned you that you need to use parentheses! Is the expression

    x^-1/2y^3/4 = (x^(-1/2))(y^(3/4)) ? Or is it

    x^-1/2y^3/4 = (x^(-1))/(2 y^(3/4)) ? It's rather hard to tell which one you mean without parentheses. If you follow strict order of operations rules, then technically you wrote the second one, which is not what you mean. So your problem, then, is

    \left(\frac{x^{-3/2}y}{x^{1/2}y^{-2}}\right)(x^{1/2}y^{-3/4}), correct?
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