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Math Help - Simplify with negetive exponents

  1. #1
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    Simplify with negetive exponents

    How des one simplify this?

    (2ab)^-2/2(ab)^3

    I am unsure if this correct,

    I removed the brackets

    -4a^-2b^-2/2a^-3b^-3

    to make exponents positive i moved to the top

    -4a^-2--3b^-2--3 +2

    which is -4ab+2

    Is this correct?
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  2. #2
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    Quote Originally Posted by kiwisa View Post
    How des one simplify this?

    (2ab)^-2/2(ab)^3

    I am unsure if this correct,

    I removed the brackets

    -4a^-2b^-2/2a^-3b^-3

    to make exponents positive i moved to the top

    -4a^-2--3b^-2--3 +2

    which is -4ab+2

    Is this correct?
    See if you follow this:

    \frac{ \left((2ab)^{-2}\right)}{\left(2(ab)^3\right)}

    =  \left(2^{-2} a^{-2} b^{-2}\right) \cdot \left(2^{-1}a^{-3}b^{-3} \right)

    =  2^{-3}a^{-5}b^{-5}

    =  \frac{1}{8a^5b^5}
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  3. #3
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    sorry on i seei have atypo in the original post the exponent should be -3 so it should read as such

    (2ab)^-2/2(ab)^-3

    But with the equation at the top where does the 2^-1 come from if it was 2^3? The rest i can understand.
    Sorry for the typo.
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  4. #4
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    Quote Originally Posted by kiwisa View Post
    sorry on i seei have atypo in the original post the exponent should be -3 so it should read as such

    (2ab)^-2/2(ab)^-3

    But with the equation at the top where does the 2^-1 come from if it was 2^3? The rest i can understand.
    Sorry for the typo.
    It's all right. For the equation I posted, you have 2(ab)^3 for the denominator. Using the law of operations the power of 3 only applies to a and b, not on the 2.

    As for the fixed equation

     \frac{(2ab)^{-2}}{2(ab)^{-3}}

    =  (2ab)^{-2} \cdot \left(2 (ab)^{-3}\right)^{-1}

    =  2^{-2}a^{-2}b^{-2} \cdot 2^{-1} (ab)^3

    =  2^{-2}a^{-2}b^{-2} \cdot 2^{-1} a^3 b^3

    =  2^{-3} ab

    =  \frac{ab}{8}
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  5. #5
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    Thanks - so to multiply the bottom instead of dividing it I put it all in brackets and multply by -1. The rest i understand.
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  6. #6
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    Hello, kiwisa!

    Simplify: . \dfrac{(2ab)^{-2}}{2(ab)^{-3}}

    We have: . \dfrac{(2ab)^{-2}}{2(ab)^{-3}} \;=\;\dfrac{2^{-2}\cdot a^{-2}\cdot b^{-2}} {2\cdot a^{-3}\cdot b^{-3}}


    Use the exponent rules: when dividing, subtract exponents.


    And we have: . 2^{-2-1}\cdot a^{-2-(-3)}\cdot b^{-1-(-3)}

    . . . . . . . . =\;2^{-3}\cdot a^1\cdot b^1

    . . . . . . . . =\; \dfrac{ab}{2^3}

    . . . . . . . . =\;\dfrac{ab}{8}
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  7. #7
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    Thanks guys got it now
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