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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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Updated on: 20 Dec 2015, 20:59
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Question Stats:

66% (00:53) correct 34% (01:16) wrong based on 151 sessions

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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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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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25 Dec 2015, 16:19
Would someone please break down the steps to solve this question, step by step?
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Posts: 5901
Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy? [#permalink]

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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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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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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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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.
Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?   [#permalink] 08 Feb 2017, 20:41
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# If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?

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