It is currently 22 Oct 2017, 23:31

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 31 Jul 2006
Posts: 440

Kudos [?]: 56 [1], given: 0

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

### Show Tags

16 Sep 2006, 09:59
1
KUDOS
5
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

69% (00:51) correct 31% (01:21) wrong based on 253 sessions

### HideShow timer Statistics

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

Attachments

x1.gif [ 1.86 KiB | Viewed 3105 times ]

Kudos [?]: 56 [1], given: 0

Senior Manager
Joined: 11 May 2006
Posts: 258

Kudos [?]: 23 [0], given: 0

### Show Tags

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

Kudos [?]: 23 [0], given: 0

SVP
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 167 [0], given: 0

### Show Tags

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

Kudos [?]: 167 [0], given: 0

Senior Manager
Joined: 31 Jul 2006
Posts: 440

Kudos [?]: 56 [0], given: 0

### Show Tags

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 ?

Kudos [?]: 56 [0], given: 0

Senior Manager
Joined: 11 May 2006
Posts: 258

Kudos [?]: 23 [0], given: 0

### Show Tags

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.

Kudos [?]: 23 [0], given: 0

Senior Manager
Joined: 28 Aug 2006
Posts: 303

Kudos [?]: 171 [1], given: 0

### Show Tags

17 Sep 2006, 05:09
1
KUDOS
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.

Kudos [?]: 171 [1], given: 0

SVP
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 167 [0], given: 0

### Show Tags

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

Kudos [?]: 167 [0], given: 0

Senior Manager
Joined: 31 Jul 2006
Posts: 440

Kudos [?]: 56 [0], given: 0

### Show Tags

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

Kudos [?]: 56 [0], given: 0

Senior Manager
Joined: 28 Aug 2006
Posts: 303

Kudos [?]: 171 [0], given: 0

### Show Tags

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.........

Kudos [?]: 171 [0], given: 0

Director
Joined: 28 Dec 2005
Posts: 750

Kudos [?]: 18 [0], given: 0

### Show Tags

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.

Kudos [?]: 18 [0], given: 0

GMAT Instructor
Joined: 04 Jul 2006
Posts: 1259

Kudos [?]: 336 [0], given: 0

### Show Tags

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.

Kudos [?]: 336 [0], given: 0

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16542

Kudos [?]: 274 [0], given: 0

### Show Tags

12 Nov 2013, 19:55
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 274 [0], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 41908

Kudos [?]: 129394 [1], given: 12197

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

### Show Tags

13 Nov 2013, 01:21
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
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
_________________

Kudos [?]: 129394 [1], given: 12197

Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?   [#permalink] 13 Nov 2013, 01:21
Display posts from previous: Sort by