# Please help index laws

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• Oct 30th 2010, 07:49 AM
stripe501
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 :)
• Oct 30th 2010, 08:01 AM
Plato
$2^{n+2}+12=4(2^n+3)$
and
$5\cdot 2^{n}+15=5(2^n+3)$
• Oct 30th 2010, 08:51 AM
stripe501
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
• Oct 30th 2010, 10:34 AM
Soroban
Hello, stripe501!

Quote:

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

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

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

Quote:

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

$\text{Answer: }\:9$

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