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# I've been lurking here for a while.. and this would be my

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Intern
Joined: 31 Oct 2006
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I've been lurking here for a while.. and this would be my [#permalink]

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04 Jan 2007, 16:26
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I've been lurking here for a while.. and this would be my first post. My apologies if it has already been done. But any help is appreciated..

If XY= 1, Then
2^(X+Y)^2
2^(X-Y)^2

equals...
answer is 16.. could you guys please show me the steps.
Much appreciated.. keep up the good work!

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Director
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04 Jan 2007, 17:57
Ignore my last reply

Last edited by ggarr on 04 Jan 2007, 18:35, edited 2 times in total.

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Intern
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04 Jan 2007, 18:17
I should have probably mentioned...
2^(X+Y)^2
2^(X-Y)^2

is 2^(X+Y)^2 / 2^(X-Y)^2
sorry for any confusion....

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Intern
Joined: 14 Jul 2005
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Location: California

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04 Jan 2007, 20:02
crxover wrote:
I should have probably mentioned...
2^(X+Y)^2
2^(X-Y)^2

is 2^(X+Y)^2 / 2^(X-Y)^2
sorry for any confusion....

2^(X+Y)^2 / 2^(X-Y)^2

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

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

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

= 2^4xy

xy = 1
so 2^4xy = 16

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Manager
Joined: 20 Dec 2004
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04 Jan 2007, 22:03
2^(X+Y)^2 / 2^(X-Y)^2
=[2^(x^2 +y^2 + 2xy)]/[2^(x^2 +y^2 - 2xy)]
Since xy=1
Hence above equation
=[2^(x^2 +y^2 + 2)]/[2^(x^2 +y^2 - 2)]
=[2^(x^2 +y^2) * 2^2)]/[2^(x^2 +y^2) * 2^- 2)]
= 2^(2+2) Note: 2^(x^2 +y^2) in numerator & denominator cancel out
=16
_________________

Regards

Subhen

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Intern
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05 Jan 2007, 08:32
Thanks guys!.. really almost seems silly posting it when you break it down so easily like that
but that's what the help is for i guess .. keep up the good work

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Senior Manager
Joined: 24 Nov 2006
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05 Jan 2007, 15:39
When you have a division of a given number with different exponents, just subtract them. In this case:

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

The exponent would be: x^2 + 2xy + y^2 - (x^2 -2xy + y^2) = 4xy = 4.

2^4 = 16.

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05 Jan 2007, 15:39
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# I've been lurking here for a while.. and this would be my

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