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(3^9-3^8)(3^7-3^6) equivilant to???

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Re: (3^9-3^8)(3^7-3^6) equivilant to???

by x2suresh » Wed Feb 11, 2009 1:01 pm
marcusking wrote:Is there a quick way to solve the following

(3^9-3^8)(3^7-3^6) =

a. 3^4
b. 3^14
c. 6^14
d. 2*3^14
e. 4*3^14
(3^9-3^8)(3^7-3^6) = 3^8(3-1)*3^6(3-1) = 4*3^14
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Re: (3^9-3^8)(3^7-3^6) equivilant to???

by marcusking » Wed Feb 11, 2009 1:09 pm
x2suresh wrote: (3^9-3^8)(3^7-3^6) = 3^8(3-1)*3^6(3-1) = 4*3^14
What property is there that allows you to do this? I'm confused as to why this works. Is this something I will just need to remember?

Any x^k-x^k-1 = (x-1)x^k-1 ??? am I understanding this correct?
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by dmateer25 » Wed Feb 11, 2009 1:17 pm
(3^9-3^8)(3^7-3^6)

Take the common factor 3^8 out of (3^9-3^8). You will be left with

3^8(3-1) = 3^8*2

Take the common factor 3^6 out of (3^7-3^6). You will be left with
3^6(3-1) = 3^6 * 2

Now when you multiply these together you add the exponents on the 3's.

You will come up with 3^14 *2^2 = 3^14 * 4
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by Stuart@KaplanGMAT » Wed Feb 11, 2009 1:53 pm
dmateer25 wrote:(3^9-3^8)(3^7-3^6)

Take the common factor 3^8 out of (3^9-3^8). You will be left with

3^8(3-1) = 3^8*2

Take the common factor 3^6 out of (3^7-3^6). You will be left with
3^6(3-1) = 3^6 * 2

Now when you multiply these together you add the exponents on the 3's.

You will come up with 3^14 *2^2 = 3^14 * 4
To show all the steps:

3^9 - 3^8

Well, 3^9 = 3^1 * 3^8, so:

3^1(3^8) - 1(3^8) = (3-1)(3^8) = 2(3^8)
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by marcusking » Thu Feb 12, 2009 5:14 am
Ahhhh! Makes perfect sense now, thanks to all.
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by Chirawat Sungkamee » Mon Aug 20, 2012 11:35 pm
SOLUTION IN DETAIL:

Due to x^a * x^b = x^a+b, therefore

(3^9 - 3^8)(3^7 - 3^6) = (3^8 * 3^1 - 3^8)(3^6 * 3^1 - 3^6)
= (3^8 * 3 - 3^8)(3^6 * 3 - 3^6)
= {3^8(3-1)}{3^6(3-1)}
= 3^8 * 2 * 3^6 * 2
= 2 * 2 * 3^8 * 3^6
= 4 * 3^8+6
= 4 * 3^14
The answer is (E).
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