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Manager  Status: Current MBA Student
Joined: 19 Nov 2009
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Concentration: Finance, General Management
GMAT 1: 720 Q49 V40
If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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Question Stats: 72% (01:14) correct 28% (01:45) wrong based on 1070 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
Math Expert V
Joined: 02 Sep 2009
Posts: 64158
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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12
13
tonebeeze 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

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

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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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56
14
WishMasterUA wrote:
if xy=1, what is the value of (2^((x+y)^2))/(2^((x-y)^2))?
1) 2
2) 4
3) 8
4) 16
5) 32

$$\frac{2^{(x+y)^2}}{2^{(x+y)^2}}$$

$$=\frac{2^{(x^2+y^2+2xy)}}{2^{(x^2+y^2-2xy)}}$$ [:Note: $$(x+y)^2=x^2+y^2+2xy \hspace{3} & \hspace{3} (x-y)^2=x^2+y^2-2xy$$]

$$=2^{(x^2+y^2+2xy-(x^2+y^2-2xy))}$$ [:Note: $$\frac{x^m}{x^n}= x^{(m-n)}$$ ]

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

$$=2^{(4xy)}$$

$$=2^{(4xy)}=2^4=16$$ [:Note: xy=1]

Ans: "D"
##### General Discussion
Manager  Status: Current MBA Student
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Posts: 89
Concentration: Finance, General Management
GMAT 1: 720 Q49 V40
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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3
1
Bunuel wrote:
tonebeeze 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

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

Hey Bunuel,

Thanks for the response. I totally understand the key logic of this problem. My method:
$$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x^2+2xy+y^2) - (x^2-2xy + y^2)} =2^{(x^2+2xy+y^2 - x^2+ 2xy - y^2)}=2^{(4xy)}=2^4=16$$

Your method looks more simple and I would like to understand it. I just got a little lost during your factoring-out transition from $$2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}$$ . Can u please explain. Thanks!
Math Expert V
Joined: 02 Sep 2009
Posts: 64158
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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5
3
tonebeeze wrote:
Bunuel wrote:
tonebeeze 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

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

Hey Bunuel,

Thanks for the response. I totally understand the key logic of this problem. My method:
$$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x^2+2xy+y^2) - (x^2-2xy + y^2)} =2^{(x^2+2xy+y^2 - x^2+ 2xy - y^2)}=2^{(4xy)}=2^4=16$$

Your method looks more simple and I would like to understand it. I just got a little lost during your factoring-out transition from $$2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}$$ . Can u please explain. Thanks!

$$a^2-b^2=(a+b)*(a-b)$$, so $$(x+y)^2-(x-y)^2=((x+y)+(x-y))*((x+y)-(x-y))=2x*2y=4xy$$.

Hope it's clear.
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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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$$2^{(x+y)^2}/2^{(x-y)^2}=2^{(x+y)^2-(x-y)^2}$$

This part is very unclear to me. I do understand the basics of negative square is just "one over", but I dont understand this. The math compendium barely touches on this either. Can someone explain this please?
Math Expert V
Joined: 02 Sep 2009
Posts: 64158
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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erikvm wrote:
$$2^{(x+y)^2}/2^{(x-y)^2}=2^{(x+y)^2-(x-y)^2}$$

This part is very unclear to me. I do understand the basics of negative square is just "one over", but I dont understand this. The math compendium barely touches on this either. Can someone explain this please?

$$\frac{a^n}{a^m}=a^{n-m}$$

Below might help:
Theory on Exponents: math-number-theory-88376.html
Tips on Exponents: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html
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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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11
1
Hi All,

Most Quant questions can be approached in a variety of ways, so it's useful to practice more than one method during your studies. In this question, it appears that all of the posters took the same Algebraic approach (which is fine), but was that approach really the fastest and easiest way to get to the solution.....?

Watch what happens when we TEST VALUES....

We're told that XY = 1. Since the answer choices are all numbers, one of them MUST be the solution to the equation, so I should be able to use ANY combination of X and Y that I choose (as long as the product of those values = 1).

Let's try...
X = 1
Y = 1

The question then becomes...what is the value of (2^4)/(2^0)?

(2^4)/(2^0) =
16/1 =
16

GMAT assassins aren't born, they're made,
Rich
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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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Bunuel wrote:
tonebeeze 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

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

Hi Bunuel,

Could you help me with the following question?

There is a rule for exponents that: $$(n^a)^b= n^{a*b}$$

So If we have:

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

Why woudn't it be the following?

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

I'm confused because if we have $$2^{x^2}$$ the value when $$x=3$$, is the value $$2^{3·2}=2^6$$ or $$2^{2^3}=2^8$$?

Math Expert V
Joined: 02 Sep 2009
Posts: 64158
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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1
guillemgc wrote:
Bunuel wrote:
tonebeeze 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

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

Hi Bunuel,

Could you help me with the following question?

There is a rule for exponents that: $$(n^a)^b= n^{a*b}$$

So If we have:

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

Why woudn't it be the following?

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

I'm confused because if we have $$2^{x^2}$$ the value when $$x=3$$, is the value $$2^{3·2}=2^6$$ or $$2^{2^3}=2^8$$?

Because it's $$2^{(x+y)^2}$$ and not $$(2^{(x+y)})^2$$

$$(a^m)^n=a^{mn}$$

$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$ (if exponentiation is indicated by stacked symbols, the rule is to work from the top down).
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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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1
tonebeeze 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

We can simplify the given expression:

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

Expanding the exponents in both the numerator and the denominator, we have:

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

We subtract the denominator’s exponent from the numerator’s exponent:

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

2^(2xy + 2xy) = 2^(4xy)

Since xy = 1, 2^4xy = 2^4 = 16.

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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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HI!

I got the right answer but with different approach...

I assumed that xy=1 if |x|=1 and |y|=1;

So if we substitute x as 1 or -1 and y as 1 or -1 both lead to 16.

Can this be a correct solution?
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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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Hi aaliyahkhalifa,

YES - your approach (TESTing VALUES) works perfectly on this question; compared with some of the longer 'math-heavy' approaches, it's considerably easier and faster. That's something to keep in mind as you continue to study for the GMAT. Most questions can be approached in more than one way - and it's possible that "your way" of dealing with a question might not actually be the most efficient option. Learning multiple approaches/Tactics can make the overall GMAT a lot easier to deal with and is essential to maximizing your performance on Test Day.

GMAT assassins aren't born, they're made,
Rich
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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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2
what the xxxx

my lord

i just took prep1 ...spent like 8 min on this question ..eventually i had to give up ,,

It was the very first question !! can you believe it ???? 8 min wasted on this one .......

what i saw is different from this lol

2(x+y)^2 / 2(x−y)^2

no wonder i couldnt get an answer ....my lord ..very disgusting

my fault ????? geeeeeeeeeeeeeeeeeee
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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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1
tonebeeze 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

$$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}$$
$$= 2^{(x+y)^2 - (x-y)^2}$$
$$= 2^{(4xy)}$$
$$= 2^{(4xy)}$$
$$= 2^4$$
$$= 16$$
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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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1
newyork2012 wrote:
what the xxxx

my lord

i just took prep1 ...spent like 8 min on this question ..eventually i had to give up ,,

It was the very first question !! can you believe it ???? 8 min wasted on this one .......

what i saw is different from this lol

2(x+y)^2 / 2(x−y)^2

no wonder i couldnt get an answer ....my lord ..very disgusting

my fault ????? geeeeeeeeeeeeeeeeeee

I did the exact same thing, I don't think that it is presented as clearly as it should be.

Rather frustrating.
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Posts: 3
If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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Bunuel wrote:
tonebeeze 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

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

Thank you for the answer but I am having a hard time understanding why when you bring this "2^{(x-y)^2" to the exponent up top, you don't bring the 2 as well?
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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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Formula to be used:

$$x^a/x^b = x^{a-b}$$

$$2^{(x+y)^2}$$ / $$2^{(x-y)^2}$$= $$2^{(x+y)^2 - (x-y)^2}$$

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

= $$2^{(4xy)}$$ = $$2^{(4*1)}$$ = $$2^4$$ = 16

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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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EMPOWERgmatRichC wrote:
Hi All,

Most Quant questions can be approached in a variety of ways, so it's useful to practice more than one method during your studies. In this question, it appears that all of the posters took the same Algebraic approach (which is fine), but was that approach really the fastest and easiest way to get to the solution.....?

Watch what happens when we TEST VALUES....

We're told that XY = 1. Since the answer choices are all numbers, one of them MUST be the solution to the equation, so I should be able to use ANY combination of X and Y that I choose (as long as the product of those values = 1).

Let's try...
X = 1
Y = 1

The question then becomes...what is the value of (2^4)/(2^0)?

(2^4)/(2^0) =
16/1 =
16

GMAT assassins aren't born, they're made,
Rich

I get the algebraic way to to do and I get number picking X=Y=1 to get 16 as well. What I don't get is why if we choose X=Y=(-1) we don't get 16.. Nor if we choose X=2 and Y=0.5. X=4 and Y=0.25.. ???
They all satisfy the original requirement of X*Y=1..
Can some one please enlighten me? Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?   [#permalink] 12 Feb 2020, 03:56

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