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
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(3^9-3^8)(3^7-3^6) equivilant to???
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marcusking
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marcusking
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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?x2suresh wrote: (3^9-3^8)(3^7-3^6) = 3^8(3-1)*3^6(3-1) = 4*3^14
Any x^k-x^k-1 = (x-1)x^k-1 ??? am I understanding this correct?
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dmateer25
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(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
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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Stuart@KaplanGMAT
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To show all the steps: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
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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marcusking
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Chirawat Sungkamee
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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).
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).
