inidices

**tim_mannire** · April 14th 2010, 06:38 PM

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)

**tim_mannire** · April 14th 2010, 06:39 PM

Originally Posted by tim_mannire

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..

**Gusbob** · April 14th 2010, 06:43 PM

Originally Posted by tim_mannire

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

**Sudharaka** · April 14th 2010, 06:43 PM

Originally Posted by tim_mannire

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!

**tim_mannire** · April 14th 2010, 06:50 PM

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

**Sudharaka** · April 14th 2010, 06:58 PM

Originally Posted by tim_mannire

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 $2^{n-1}$ ?

**tim_mannire** · April 14th 2010, 07:02 PM

Originally Posted by Sudharaka

Dear tim_mannire,

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

3 + 5 (2^n)

**Sudharaka** · April 14th 2010, 07:13 PM

Originally Posted by tim_mannire

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?

**tim_mannire** · April 14th 2010, 07:19 PM

Originally Posted by Sudharaka

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

**Sudharaka** · April 14th 2010, 07:29 PM

Originally Posted by tim_mannire

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.

**tim_mannire** · April 14th 2010, 07:49 PM

Originally Posted by Sudharaka

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

**Sudharaka** · April 14th 2010, 08:01 PM

Originally Posted by tim_mannire

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.

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