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If xy = 1

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by babachal » Thu Jun 12, 2008 3:23 pm
Hey, can you give us full question?
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by netigen » Thu Jun 12, 2008 3:38 pm
Question is what is the value of

2^(x + Y)^2)/2^(x - Y)^2
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by egybs » Thu Jun 12, 2008 4:02 pm
ahhhh.. in that case it's easy!

2^(x+y)^2 / 2^(x-y)^2

2^((x+y)^2 - (x-y)^2)

2^(x^2 + 2xy + y^2 - x^2 + 2xy - y^2)

2^(4xy)

which as we know is equal to 2^4... so 16.
Last edited by egybs on Thu Jun 12, 2008 4:21 pm, edited 1 time in total.
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by babachal » Thu Jun 12, 2008 4:17 pm
egybs, you did mistake in the last step

2^(x^2+2xy+y^2-x^2+2xy-y^2)+2^4xy=2^4=16
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by egybs » Thu Jun 12, 2008 4:20 pm
oops! yep. i did.
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by s_raizada » Fri Jun 13, 2008 1:36 am
Hi guys,
question is not what you are calculating. See the screenshot of the question and it is form GMAT prep
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mathquestion.JPG
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by AleksandrM » Fri Jun 13, 2008 9:19 am
Undifined or 25/9????? :?
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by egybs » Fri Jun 13, 2008 9:32 am
Can you give us the answer options...
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by Ian Stewart » Fri Jun 13, 2008 11:23 am
As has been posted elsewhere, there's a typo in the question itself. The solution posted above by netigen, with the x and y terms in the exponents, presented the question in the way it's supposed to be written.
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by GMAT dreamer » Tue Nov 18, 2008 12:16 pm
We can make things simpler...
If xy=1; it tells that x=1 and y=1, right?

so, 2^(1+1)^2/2^(1-1)^2=
2^2^2/2^0^2=
2^4/2^0=
16/1= 16

Very simple we put numbers.
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by cramya » Tue Nov 18, 2008 3:40 pm
If xy=1; it tells that x=1 and y=1, right?
In general if xy=1 does not necessarily mean x=1 and y=1 unless there are other properties about x and y also given that confirms the fact that x=1 y=1

To quote a few:

x=-1 y=-1 xy=1

x=5 y=1/5 xy =1

x=-5 y=-1/5 xy=1

Unless it says x and y are positive integers and xy=1 we cannot assume
x=1 and y=1

Hope this helps!
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by rochy81 » Sun Aug 16, 2009 1:02 pm
Ahh, I was just stuck on this. I'm kicking myself...
Anyone happen to know what the difficulty of this question is?
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