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help...

by shibal » Sun Aug 29, 2010 2:36 pm
I just took an official CAT with a pretty good outcome... but I don't know why I got the FIRST freakin question wrong.... i did 75 times and still can't get to the right answer...

(2^(4-1)^2)/(2^(3-2)

OA: [spoiler]2^8[/spoiler]

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by Rahul@gurome » Sun Aug 29, 2010 2:57 pm
shibal wrote:I just took an official CAT with a pretty good outcome... but I don't know why I got the FIRST freakin question wrong.... i did 75 times and still can't get to the right answer...

(2^(4-1)^2)/(2^(3-2)

OA: [spoiler]2^8[/spoiler]
(2^(4-1)^2)/(2^(3-2) = (2^3^2)/(2^1) = (2^9)/2 = 2^(9-1) = 2^8
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by shibal » Sun Aug 29, 2010 3:08 pm
Rahul@gurome wrote:
shibal wrote:I just took an official CAT with a pretty good outcome... but I don't know why I got the FIRST freakin question wrong.... i did 75 times and still can't get to the right answer...

(2^(4-1)^2)/(2^(3-2)

OA: [spoiler]2^8[/spoiler]
(2^(4-1)^2)/(2^(3-2) = (2^3^2)/(2^1) = (2^9)/2 = 2^(9-1) = 2^8
but by exponent rules (2^3)^2 be 2^(3*2)???

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by Rahul@gurome » Sun Aug 29, 2010 3:35 pm
shibal wrote:
but by exponent rules (2^3)^2 be 2^(3*2)???
The question is (2^(4-1)^2)/(2^(3-2)

Numerator = (2^(4-1)^2) = 2^3^2
3^2 = 9, so numerator reduces to 2^9
Now denominator = 2^(3-2) = 2^1 = 2
Therefore, 2^9/2 = 2^(9-1) = 2^8

We cannot write (2^3)^2 = 2^(3*2), as 2^3^2 = 2^(3^2) = 2^9
2^3^2 means 2 is a power of 3, which is again a power of 2, so first solve 3^2 = 9 and then the expression reduces to 2^9.
Does that help?
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by shibal » Sun Aug 29, 2010 3:59 pm
Rahul@gurome wrote:
shibal wrote:
but by exponent rules (2^3)^2 be 2^(3*2)???
The question is (2^(4-1)^2)/(2^(3-2)

Numerator = (2^(4-1)^2) = 2^3^2
3^2 = 9, so numerator reduces to 2^9
Now denominator = 2^(3-2) = 2^1 = 2
Therefore, 2^9/2 = 2^(9-1) = 2^8

We cannot write (2^3)^2 = 2^(3*2), as 2^3^2 = 2^(3^2) = 2^9
2^3^2 means 2 is a power of 3, which is again a power of 2, so first solve 3^2 = 9 and then the expression reduces to 2^9.
Does that help?

hummm so can I conclude that (2^3)^2 is different from (2^3^2)?

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by Rahul@gurome » Sun Aug 29, 2010 4:28 pm
shibal wrote:
hummm so can I conclude that (2^3)^2 is different from (2^3^2)?
Yes definitely, they are different.
2^3 = 8 so (2^3)^2 = 8^2 = 64
and 2^3^2 = 2^(3^2) = 2^9
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