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# If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?

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Senior Manager
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If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]

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16 Sep 2006, 09:59
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69% (01:49) correct 31% (01:21) wrong based on 250 sessions

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If xy = 1, what is the value of $$\frac {2^{(x + y)^2}} {2^{(x - y)^2}$$ ?

A. 2
B. 4
C. 8
D. 16
E. 32

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-xy-1-what-is-the-value-of-2-x-y-2-2-x-y-107238.html
[Reveal] Spoiler: OA

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16 Sep 2006, 10:06
= 2^[(x+y)^2 - (x-y)^2]
= 2^[4xy]
= 2^4 since xy = 1
= 16
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16 Sep 2006, 10:07
U can expand 2^((x+y)^2) :
2^((x+y)^2)
= 2^(x^2+2*x*y+y^2)
= 2^(x^2+y^2) * 2^(2*x*y)

Similarly,
2^((x-y)^2)
= 2^(x^2+y^2) * 2^(-2*x*y)

Since x*y=1

2^((x+y)^2) = 2^(x^2+y^2) * 2^(2) = 4*2^(x^2+y^2)
and
2^((x-y)^2) = 2^(x^2+y^2) * 2^(-2) = 2^(x^2+y^2) /4

Hence,
2^((x+y)^2) / 2^((x-y)^2)
= 4 * 2^(x^2+y^2) / ( 2^(x^2+y^2) /4 )
= 4 / (1/4)
= 16
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16 Sep 2006, 10:18
iced_tea wrote:
= 2^[(x+y)^2 - (x-y)^2]
= 2^[4xy]
= 2^4 since xy = 1
= 16

so when you bring 2^[(x-y)^2] up you changed that sign of (x-y) (that I understand), but shouldn't you change the sign of ^2 as well?

for ex: 2^-(x-y)^-2 ?
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16 Sep 2006, 10:22
Nsentra wrote:
iced_tea wrote:
= 2^[(x+y)^2 - (x-y)^2]
= 2^[4xy]
= 2^4 since xy = 1
= 16

so when you bring 2^[(x-y)^2] up you changed that sign of (x-y) (that I understand), but shouldn't you change the sign of ^2 as well?

for ex: 2^-(x-y)^-2 ?

no ...the property to use here is

a^b/a^c = a ^(b-c)

consider b = (x+y)^2 and c = (x-y)^2 here.
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17 Sep 2006, 05:09
1
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Folks, I guess this is a simple one from algebra.

To be more fast I think we can do this
Since XY = 1 the answer for the given question must be same for any values of X and Y that satisfy XY = 1

So take X=1 and Y=1.
The given value becomes 2^4 = 16
And I think the problem is killed.

What do u say?

Regards,
_________________

Last edited by cicerone on 25 Sep 2008, 01:00, edited 1 time in total.
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17 Sep 2006, 08:06
cicerone wrote:
Folks, I guess this is a simple one from algebra.

To be more fast I think we can do this
Since XY = 1 the answer for the given question must be same for any values of X and Y that satisfy XY = 1

So take X=1 and Y=1.
The given value becomes 2^4 = 16
And I think the problem is killed.

What do u say?

Regards,

It's a good shortcut ... Probably able to make us preserve our precious energy for other questions
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17 Sep 2006, 08:13
cicerone wrote:
Folks, I guess this is a simple one from algebra.

To be more fast I think we can do this
Since XY = 1 the answer for the given question must be same for any values of X and Y that satisfy XY = 1

So take X=1 and Y=1.
The given value becomes 2^4 = 16
And I think the problem is killed.

What do u say?

Regards,

that's a bit dangerous, no? x=-2 and y=-1/2 could possibly give us one of the answers if the test makes wanted to catch this
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17 Sep 2006, 21:59
Nsentra wrote:
cicerone wrote:
Folks, I guess this is a simple one from algebra.

To be more fast I think we can do this
Since XY = 1 the answer for the given question must be same for any values of X and Y that satisfy XY = 1

So take X=1 and Y=1.
The given value becomes 2^4 = 16
And I think the problem is killed.

What do u say?

Regards,

that's a bit dangerous, no? x=-2 and y=-1/2 could possibly give us one of the answers if the test makes wanted to catch this

Friend it's not at all dangerous.
For any given value of X and Y that satisfy XY=1, the given expression always gives only one answer. Try to understand this.........
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18 Sep 2006, 00:17
cicerone wrote:
Nsentra wrote:
cicerone wrote:
Folks, I guess this is a simple one from algebra.

To be more fast I think we can do this
Since XY = 1 the answer for the given question must be same for any values of X and Y that satisfy XY = 1

So take X=1 and Y=1.
The given value becomes 2^4 = 16
And I think the problem is killed.

What do u say?

Regards,

that's a bit dangerous, no? x=-2 and y=-1/2 could possibly give us one of the answers if the test makes wanted to catch this

Friend it's not at all dangerous.
For any given value of X and Y that satisfy XY=1, the given expression always gives only one answer. Try to understand this.........

it just so happens that in the equation presented, x=1 and y=1 works. I don't agree that it will always work.....what if you had a term like x/y?
by your logic, x/y would be reduced to 1, which is clearly overstepping the given data.
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18 Sep 2006, 00:38
cicerone wrote:
Folks, I guess this is a simple one from algebra.

To be more fast I think we can do this
Since XY = 1 the answer for the given question must be same for any values of X and Y that satisfy XY = 1

So take X=1 and Y=1.
The given value becomes 2^4 = 16
And I think the problem is killed.

What do u say?

Regards,

Very good point! By the wording of the question, any values of x and y whose product is 1 will yield the same value of 2^.../2^... Here's a great question for picking numbers, though the algebraic approach is quite manageable.
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12 Nov 2013, 19:55
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Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]

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13 Nov 2013, 01:21
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Nsentra wrote:
If xy = 1, what is the value of $$\frac {2^{(x + y)^2}} {2^{(x - y)^2}$$ ?

A. 2
B. 4
C. 8
D. 16
E. 32

$$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-xy-1-what-is-the-value-of-2-x-y-2-2-x-y-107238.html
_________________
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?   [#permalink] 13 Nov 2013, 01:21
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