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# [3^-(x+y)]/[3^-(x-y)]=?

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Senior Manager
Joined: 08 Aug 2005
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05 May 2006, 23:33
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[3^-(x+y)]/[3^-(x-y)]=?

1) x=2
2) y=3
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Joined: 29 Dec 2005
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11 May 2006, 12:02
getzgetzu wrote:
[3^-(x+y)]/[3^-(x-y)]=?

1) x=2
2) y=3

i guess i did it recently but i find it very difficult to serch past posting under math section where as under verbal section it easier to do so..

= 3^-(x+y)]/[3^-(x-y)]
= [3^(x-y)]/ [3^(x+y)]
= (3^x) 3^(-y)/ (3^x 3^y)
= 1/[(3^y) (3^y)]
= 1/3^2y

statement 2 is suff... so B.
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11 May 2006, 12:55
The answer is B. When you simplify, you are left with

(3^0)*(3^(-2y))

1/(3^2y)
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11 May 2006, 13:29
I got B by simplifying as follows:

[3^-(x+y)]/[3^-(x-y)]
[3^(-x-y)]/[3^(-x+y)]
[3^-x * 3^-y]/[3^-x * 3^y]
3^-y / 3^y
3^-2y
11 May 2006, 13:29
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