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Math Help - exponents

  1. #1
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    exponents

    Could someone help me out with these?

    Using the laws of indices to simplify each of the following, giving answers with positive indices only.

    1. 3^(3x +4).18^(-x-2)/24^(x+3).4^(-2x-1)



    2. 21^(3x+1).7(2-x)/49^(x-1).27^(x-4)
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  2. #2
    Senior Member tukeywilliams's Avatar
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    Quote Originally Posted by Dean View Post
    Could someone help me out with these?

    Using the laws of indices to simplify each of the following, giving answers with positive indices only.

    1. 3^(3x +4).18^(-x-2)/24^(x+3).4^(-2x-1)



    2. 21^(3x+1).7(2-x)/49^(x-1).27^(x-4)
    1.  \frac{3^{3x+4} \cdot 18^{-x-2}}{24^{x+3} \cdot 4^{-2x-1}} = \frac{3^{3x+4} \cdot 3^{-x-2} \cdot 3^{-x-2} \cdot 2^{-x-2}}{2^{x+3} \cdot 2^{x+3} \cdot 3^{x+3} \cdot 2^{x+3} \cdot 2^{-2x-1} \cdot 2^{-2x-1}} = \frac{3^{x}}{2^{9} \cdot 3^{x+3}} = \frac{1}{3^{3} \cdot 2^{9}}





    2.  \frac{21^{3x+1} \cdot 7^{2-x}}{49^{x-1} \cdot 27^{x-4}} = \frac{7^{3x+1} \cdot 3^{3x+1} \cdot 7^{2-x}}{7^{x-1} \cdot 7^{x-1} \cdot 27^{x-4}} = \frac{7^{2x+3} \cdot 3^{3x+1}}{7^{2x-2} \cdot 3^{x-4} \cdot 3^{x-4} \cdot 3^{x-4}} = \frac{7^{5} \cdot 3^{2x+5}}{3^{2x-8}} = 7^{5} \cdot 3^{13}
    Last edited by tukeywilliams; August 1st 2007 at 12:57 AM.
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  3. #3
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    Quote Originally Posted by Dean View Post
    Could someone help me out with these?
    Using the laws of indices to simplify each of the following, giving answers with positive indices only.

    1. 3^(3x +4).18^(-x-2)/24^(x+3).4^(-2x-1)
    ...
    Hello,

    1. I assume that you mean exponents and not indices (?)
    2. I assume that you are familiar with the laws and rules to calculate powers.
    3. I assume that you want to simplify

    \frac{3^{3x+4} \cdot 18^{-x-2}}{24^{x+3} \cdot 4^{-2x-1}}

    Factor all bases and transform to a produkt:

    \frac{3^{3x+4} \cdot 18^{-x-2}}{24^{x+3} \cdot 4^{-2x-1}} = 3^{3x} \cdot 3^4 \cdot 3^{-2x} \cdot 2^{-x} \cdot 3^{-4} \cdot 2^{-2} \cdot 3^{-x} \cdot 2^{-3x} \cdot 3^{-3} \cdot 2^{-9} \cdot 2^{4x} \cdot 2^2 = 3^0 \cdot 2^0 \cdot 3^{-3} \cdot 2^{-9} = \frac{1}{24^3}
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  4. #4
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    Which way is correct?
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  5. #5
    Senior Member tukeywilliams's Avatar
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     \frac{1}{24^3} = \frac{1}{3^{3} \cdot 2^{9}} . Both ways are correct. I just converted the  18^{-x-2} to  (3 \cdot 3 \cdot 2)^{-x-2} and worked from there. Same with the second problem.
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  6. #6
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    Oh, thanks so much guys.
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