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Thread: Please help index laws

  1. October 30th 2010, 07:49 AM #1
    stripe501
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    Please help index laws

    Hi, in my math textbook I have three question which the textbook does not teach you how to simplify please help. I don't just want the answer I want to know how to get the answer again if I receive a similar question.

    These are the three questions. The correct answers are below each question question

    Simplify...

    131. (2^(n+2) + 12) / (5 * 2^n + 15)
    4/5
    132. (2^(2n+3) - (2n)^2) / 2^n
    7 * (2^n)
    133. (3^(n+1) + 9) / (3^(n-1) + 1)
    9

    Thanks in advance
    Last edited by stripe501; October 30th 2010 at 08:39 AM.
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  2. October 30th 2010, 08:01 AM #2
    Plato
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    2^{n+2}+12=4(2^n+3)
    and
    5\cdot 2^{n}+15=5(2^n+3)
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  3. October 30th 2010, 08:51 AM #3
    stripe501
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    Thankyou for your quick response.
    But I still need help with the other 2 questions if you could
    I must be doing something wrong :S
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  4. October 30th 2010, 10:34 AM #4
    Soroban
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    Hello, stripe501!

    132.\;\;\dfrac{2^{2n+3} - (2n)^2}{2^n}

    \text{Answer: }\:7\cdot2^n

    We have: . \displaystyle \frac{2^{2n+3} - 2^{2n}}{2^n} \;=\;\frac{2^{2n}(2^3-1)}{2^n} \;=\;\frac{2^{2n}\cdot 7}{2^n} \;=\;7\cdot2^n



    133.\;\;\dfrac{3^{n+1} + 9}{3^{n-1} + 1}

    \text{Answer: }\:9

    \displaystyle \text{We have: }\;\frac{3^{n+1} + 3^2}{3^{n-1}+1} \;=\;\frac{3^2(3^{n-1}+1)}{3^{n-1}+1} \;=\;3^2 \;=\;9
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  5. October 30th 2010, 08:50 PM #5
    stripe501
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    Both of you thanks for your help. Much appreciated
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  6. October 31st 2010, 12:26 AM #6
    stripe501
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    I under stand everything accept for how you get from (2^(2n) * 7) / 2n
    to 7 * 2^n
    could you please explain
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  7. October 31st 2010, 12:30 AM #7
    Educated
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    \dfrac{2^{2n} \cdot 7}{2^n} = \dfrac{(2^n)^2\cdot 7}{2^n} = \dfrac{2^n \cdot 2^n \cdot 7}{2^n} = 2^n \cdot 7
    Thanks from stripe501
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  8. October 31st 2010, 12:37 AM #8
    stripe501
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    Ahhh, I see now. Thank you very much
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