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Math Help - Index laws

  1. #1
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    Index laws

    Hello

    Here's the problem I need a hand with, I need to simplify it:

    (6^2n-3^n)/(12^n-1)

    Thanks in advance, all help is really appreciated
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  2. #2
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    Quote Originally Posted by listeningintently View Post
    Hello

    Here's the problem I need a hand with, I need to simplify it:

    (6^2n-3^n)/(12^n-1)

    Thanks in advance, all help is really appreciated
    1. Expand the brackets:

    \dfrac{6^{2n}-3^n}{12^{n-1}} = \dfrac{(6^2)^n}{\frac1{12} \cdot 12^n} - \dfrac{3^n}{\frac1{12} \cdot 12^n} =  \dfrac{12 \cdot 36^n}{ 12^n} - \dfrac{12 \cdot 3^n}{3^n \cdot 4^n} = 12 \cdot \left(3^n - \dfrac1{4^n}  \right)
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  3. #3
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    Hello, listeningintently!

    Simplify: . \frac{6^{2n} - 3^n}{12^{n-1}}

    We have: . \frac{(2\!\cdot\!3)^{2n} - 3^n}{(2^2\!\cdot\!3)^{n-1}} \;=\;\frac{2^{2n}\!\cdot\!3^{2n} - 3^n}{2^{2(n-1)}\!\cdot\!3^{n-1}}

    Factor: . \frac{3^n\left(2^{2n}\!\cdot\!3^n - 1\right)}{3^{n-1}\!\cdot\!2^{2n-2}} \;=\;\frac{3\left(2^{2n}\!\cdot\!3^n - 1\right)}{2^{2n-2}}


    There is no end to the "simplifying" . . .

    3\left(\frac{2^{2n}\!\cdot\!3^n}{2^{2n-2}} - \frac{1}{2^{2n-2}}\right)  \;=\;3\left(2^2\!\cdot\!3^n - 2^{2-2n}\right)

    Factor: . 3\!\cdot\!2^2\left(3^n - 2^{-2n}\right) \;=\; 12\left(3^n - 2^{-2n}\right) \quad\hdots\quad\text{etc.}

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