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getzgetzu
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[3^-(x+y)]/[3^-(x-y)]=? 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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Professor
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[3^-(x+y)]/[3^-(x-y)]=? 11 May 2006, 12:02
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getzgetzu
[3^-(x+y)]/[3^-(x-y)]=?

1) x=2
2) y=3
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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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gidimba
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[3^-(x+y)]/[3^-(x-y)]=? 11 May 2006, 12:55
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The answer is B. When you simplify, you are left with

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

1/(3^2y)
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kook44
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[3^-(x+y)]/[3^-(x-y)]=? 11 May 2006, 13:29
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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



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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