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

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Senior Manager
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[3^-(x+y)]/[3^-(x-y)]=? [#permalink] New post 05 May 2006, 22:33
00:00
A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
[3^-(x+y)]/[3^-(x-y)]=?

1) x=2
2) y=3
VP
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Re: DS Exponents [#permalink] New post 11 May 2006, 11:02
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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 [#permalink] New post 11 May 2006, 11:55
The answer is B. When you simplify, you are left with

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

1/(3^2y)
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 [#permalink] New post 11 May 2006, 12:29
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
  [#permalink] 11 May 2006, 12:29
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