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Math Help - inidices

  1. #1
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    inidices

    I'm having diffulculty solving this indices question.. Can anyone show me how its done.

    3 (
    2^n+1) + 5 (2^n)
    ---------------------------

    7 (2^n) - 3 (2^n-1)
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    Quote Originally Posted by tim_mannire View Post
    I'm having diffulculty solving this indices question.. Can anyone show me how its done.


    3 (2^n+1) + 5 (2^n)
    ---------------------------

    7 (2^n) - 3 (2^n-1)
    the 3( should be level with the rest of the question..
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    Quote Originally Posted by tim_mannire View Post
    I'm having diffulculty solving this indices question.. Can anyone show me how its done.

    3 (
    2^n+1) + 5 (2^n)
    ---------------------------

    7 (2^n) - 3 (2^n-1)
    Try factorising both the numerator and denominator by 2^n
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    Quote Originally Posted by tim_mannire View Post
    the 3( should be level with the rest of the question..
    Dear tim_mannire,

    Do you mean \frac{3(2^{n+1})+5(2^n)}{7(2^n)-3(2^{n-1})} ?? If it is so first you could divide both the numerator and the denominator by 2^{n-1}. Hope you could continue from here!
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    i've tried both strategies, but I really don't understand . A worked example to simplify would be good so I can work through it and understand it. thanks
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    Quote Originally Posted by tim_mannire View Post
    i've tried both strategies, but I really don't understand . A worked example to simplify would be good so I can work through it and understand it. thanks
    Dear tim_mannire,

    Can you tell me what do you get when you divide the numerator by 2^{n-1} ?
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    Quote Originally Posted by Sudharaka View Post
    Dear tim_mannire,

    Can you tell me what do you get when you divide the numerator by 2^{n-1} ?


    3 + 5 (2^n)
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    Quote Originally Posted by tim_mannire View Post
    3 + 5 (2^n)
    Dear tim_mannire,

    Incorrect. Now, \frac{2^{n+1}}{2^{n-1}}=\frac{2^{2}\times2^{n-1}}{2^{n-1}}=4 and \frac{2^n}{2^{n-1}}=\frac{2\times2^{n-1}}{2^{n-1}}=2 Did you understand? Now can you tell me the answer?
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  9. #9
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    Quote Originally Posted by Sudharaka View Post
    Dear tim_mannire,

    Incorrect. Now, \frac{2^{n+1}}{2^{n-1}}=\frac{2^{2}\times2^{n-1}}{2^{n-1}}=4 and \frac{2^n}{2^{n-1}}=\frac{2\times2^{n-1}}{2^{n-1}}=2 Did you understand? Now can you tell me the answer?

    I understand it but am struggling to apply it to the question
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  10. #10
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    Quote Originally Posted by tim_mannire View Post
    I understand it but am struggling to apply it to the question
    Dear tim_mannire,

    \frac{3(2^{n+1})+5(2^n)}{7(2^n)-3(2^{n-1})}=\frac{3\left(\frac{2^{n+1}}{2^{n-1}}\right)+5\left(\frac{2^n}{2^{n-1}}\right)}{7\left(\frac{2^n}{2^{n-1}}\right)-3\left(\frac{2^{n-1}}{2^{n-1}}\right)} Do you understand upto this? If you understand the way of dividing which I showed in the previous thread you should be able to simplify this.
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  11. #11
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    Quote Originally Posted by Sudharaka View Post
    Dear tim_mannire,

    \frac{3(2^{n+1})+5(2^n)}{7(2^n)-3(2^{n-1})}=\frac{3\left(\frac{2^{n+1}}{2^{n-1}}\right)+5\left(\frac{2^n}{2^{n-1}}\right)}{7\left(\frac{2^n}{2^{n-1}}\right)-3\left(\frac{2^{n-1}}{2^{n-1}}\right)} Do you understand upto this? If you understand the way of dividing which I showed in the previous thread you should be able to simplify this.

    I tried what you said and i came up with 53/23.. but i also got 22/17 with a different way.. do any of them figures sound correct.. if not what is the correct final answer so i can understand how to get it. thanks
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  12. #12
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    Quote Originally Posted by tim_mannire View Post
    I tried what you said and i came up with 53/23.. but i also got 22/17 with a different way.. do any of them figures sound correct.. if not what is the correct final answer so i can understand how to get it. thanks
    Dear tim_mannire,

    The correct answer is, \frac{22}{11}=2. I think you ought to revise laws of indices.

    <br />
\frac{3(2^{n+1})+5(2^n)}{7(2^n)-3(2^{n-1})}=\frac{3\left(\frac{2^{n+1}}{2^{n-1}}\right)+5\left(\frac{2^n}{2^{n-1}}\right)}{7\left(\frac{2^n}{2^{n-1}}\right)-3\left(\frac{2^{n-1}}{2^{n-1}}\right)}<br />
=\frac{3\left(\frac{2^2\times2^{n-1}}{2^{n-1}}\right)+5\left(\frac{2\times2^{n-1}}{2^{n-1}}\right)}{7\left(\frac{2\times2^{n-1}}{2^{n-1}}\right)-3\left(\frac{2^{n-1}}{2^{n-1}}\right)}

    =\frac{(3\times2^2)+(5\times2)}{(7\times2)-(3\times1)}=\frac{22}{11}=2

    Hope this will help you.
    Last edited by Sudharaka; April 14th 2010 at 08:13 PM.
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