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Thread: Simplifying fractions with exponents

  1. November 2nd 2006, 01:36 PM #1
    violinist
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    Simplifying fractions with exponents

    It should be simple, but I keep messing it up. I really need a step-by-step description of the process if it is possible, as the problem lies in my messing up one of the steps, I imagine, but I won't know which one(s) unless I see the correct steps.

    Now, for the problem.
    (-2xyz)-³
    (x²z-³)-³

    The instructions are "Simplify".

    Thanks sooooo much for any help!

    P.S., I am homeschooled, and have the answer key. After checking the answer I realized that I have the numbers for the exponents correct, but I am somehow doing the coefficients wrong. Somewhere they are getting an "8x to the blah blah", and I have no idea how they got an 8?
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  2. November 2nd 2006, 01:51 PM #2
    CaptainBlack
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    Quote Originally Posted by violinist View Post
    It should be simple, but I keep messing it up. I really need a step-by-step description of the process if it is possible, as the problem lies in my messing up one of the steps, I imagine, but I won't know which one(s) unless I see the correct steps.


    Now, for the problem.
    (-2xyz)-³
    (x²z-³)-³

    The instructions are "Simplify".

    Thanks sooooo much for any help!

    P.S., I am homeschooled, and have the answer key. After checking the answer I realized that I have the numbers for the exponents correct, but I am somehow doing the coefficients wrong. Somewhere they are getting an "8x to the blah blah", and I have no idea how they got an 8?
    I will focus on the numerical coefficient

    We have:

    <br /> \frac{(-2x y z)^{-3}}{(x^2z^{-3})^{-3}}<br /> =(-2)^{-3}\frac{(x y z)^{-3}}{(x^2z^{-3})^{-3}}<br /> =\frac{1}{(-2)^3}\frac{(x y z)^{-3}}{(x^2z^{-3})^{-3}}<br />

    Now (-2)^3=(-1)^3\ 2^3=-(2^3)=-8 hence:

    <br /> \frac{(-2x y z)^{-3}}{(x^2z^{-3})^{-3}}<br /> =-\frac{1}{8}\ \frac{(x y z)^{-3}}{(x^2z^{-3})^{-3}}<br />

    RonL
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