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

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Senior Manager
Joined: 08 Aug 2005
Posts: 251
Followers: 1

Kudos [?]: 17 [0], given: 0

[3^-(x+y)]/[3^-(x-y)]=? [#permalink]  05 May 2006, 22:33
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(N/A)

Question Stats:

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
Joined: 29 Dec 2005
Posts: 1349
Followers: 7

Kudos [?]: 30 [0], given: 0

Re: DS Exponents [#permalink]  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.
Manager
Joined: 11 Oct 2005
Posts: 94
Followers: 1

Kudos [?]: 4 [0], given: 0

The answer is B. When you simplify, you are left with

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

1/(3^2y)
Manager
Joined: 23 Jan 2006
Posts: 192
Followers: 1

Kudos [?]: 10 [0], given: 0

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
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