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If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?

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If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?  [#permalink]

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New post Updated on: 20 Dec 2015, 20:59
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If \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2\), what is the value of xy?

(A) -1/4
(B) 1/2
(C) 0
(D) 1/4
(E) 1/2

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Originally posted by NoHalfMeasures on 20 Dec 2015, 20:25.
Last edited by Bunuel on 20 Dec 2015, 20:59, edited 1 time in total.
Edited the question.
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Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?  [#permalink]

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New post 20 Dec 2015, 20:27
2^{(x+y)^2} / 2^{(x-y)^2} = 2
=>(x+y)^2 - (x-y)^2 = 1
=> 4xy = 1
xy=1/4

Ans D
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Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?  [#permalink]

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New post 25 Dec 2015, 16:19
Would someone please break down the steps to solve this question, step by step?
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Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?  [#permalink]

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New post 27 Dec 2015, 20:43
Anonamy wrote:
Would someone please break down the steps to solve this question, step by step?



Hi,
the Q is
If \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2\), what is the value of xy?

(A) -1/4
(B) 1/2
(C) 0
(D) 1/4
(E) 1/2..

\(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2\)

the power can be added if multiplied and can be subtracted if divided..
when you take a power to numerator from denominator, you have to just add - sign..
example 1/2^2=2^(-2)..
so \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2\) will become..
\(2^{(x+y)^2-(x-y)^2} = 2\)..
\(2^{(x^2+y^2+2xy-x^2-y^2+2xy)}=2\)..
so\(2^(4xy)=2^1\).
4xy=1 or xy=1/4..
D
Hope it is clear now..
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Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?  [#permalink]

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New post 30 Jan 2017, 09:04
NoHalfMeasures wrote:
If \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2\), what is the value of xy?

(A) -1/4
(B) 1/2
(C) 0
(D) 1/4
(E) 1/2


\(2^{(x^2 +y^2 + 2xy) - (x^2 +y^2 - 2xy)}= 2^1\)

Or, \(2^{x^2 +y^2 + 2xy - x^2 - y^2 + 2xy}= 2^1\)

Or, \(2^{4xy}= 2^1\)

Or, \(4xy = 1\)

So, \(xy = \frac{1}{4}\)

Thus, answer will be (D) \(\frac{1}{4}\) ...

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Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?  [#permalink]

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New post 31 Jan 2017, 16:10
On solving
2^4xy = 2^1
4xy = 1
xy = 1/4
D
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Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?  [#permalink]

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New post 08 Feb 2017, 20:41
Have a query: If base & exponent are the same, then the exponents have to be added? Because I came across this rule, a^n.b^n=(ab)^n and applying this in the subjected problem, gave the wrong answer.
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Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?  [#permalink]

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Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?   [#permalink] 24 Dec 2018, 11:28
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