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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- Apr 14th 2010, 06:38 PM #1

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- Apr 14th 2010, 06:39 PM #2

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- Apr 14th 2010, 06:50 PM #5

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- Apr 14th 2010, 07:02 PM #7

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- Apr 14th 2010, 07:13 PM #8

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- Apr 14th 2010, 07:29 PM #10

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Dear tim_mannire,

$\displaystyle \frac{3(2^{n+1})+5(2^n)}{7(2^n)-3(2^{n-1})}=\frac{3\left(\frac{2^{n+1}}{2^{n-1}}\right)+5\left(\frac{2^n}{2^{n-1}}\right)}{7\left(\frac{2^n}{2^{n-1}}\right)-3\left(\frac{2^{n-1}}{2^{n-1}}\right)}$ 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.

- Apr 14th 2010, 07:49 PM #11

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- Apr 14th 2010, 08:01 PM #12

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Dear tim_mannire,

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

$\displaystyle

\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)}

=\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)}$

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

Hope this will help you.