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

tonebeeze

Joined: 19 Nov 2009

Last visit: 15 Jan 2015

Posts: 79

Own Kudos:

2,977

[6]

Given Kudos: 210

Status:Current MBA Student

Concentration: Finance, General Management

Schools: Tuck '15 (A) CBS '15 (A) Wharton '15 (M) Booth '15 (A) Haas '15 (A) Stanford '15 (D)

GMAT 1: 720 Q49 V40 Verified score

Schools: Tuck '15 (A) CBS '15 (A) Wharton '15 (M) Booth '15 (A) Haas '15 (A) Stanford '15 (D)

GMAT 1: 720 Q49 V40 Verified score

Posts: 79

Kudos: 2,977

[6]

02 Jan 2011, 17:12

5

Kudos

1

Bookmarks

Bunuel

tonebeeze

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

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

Answer: D.

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!

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 19 May 2025

Posts: 101,532

Own Kudos:

725,560

[18]

Given Kudos: 93,558

Products:

Expert

Expert reply

Posts: 101,532

Kudos: 725,560

[18]

02 Jan 2011, 17:20

12

Kudos

6

Bookmarks

tonebeeze

Bunuel

tonebeeze

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

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

Answer: D.

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.

fluke

Retired Moderator

Joined: 20 Dec 2010

Last visit: 24 Oct 2013

Posts: 1,104

Own Kudos:

4,967

[128]

Given Kudos: 376

Posts: 1,104

Kudos: 4,967

[128]

18 Jun 2011, 07:44

98

Kudos

29

Bookmarks

WishMasterUA

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"

erikvm

Joined: 26 Feb 2015

Last visit: 19 Mar 2017

Posts: 92

Own Kudos:

223

Given Kudos: 43

Posts: 92

Kudos: 223

04 Mar 2015, 09:36

Kudos

Bookmarks

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

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?

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 19 May 2025

Posts: 101,532

Own Kudos:

725,560

[2]

Given Kudos: 93,558

Products:

Expert

Expert reply

Posts: 101,532

Kudos: 725,560

[2]

04 Mar 2015, 09:50

1

Kudos

1

Bookmarks

erikvm

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

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

guillemgc

Joined: 26 Apr 2017

Last visit: 30 Mar 2018

Posts: 4

Own Kudos:

3

[1]

Given Kudos: 15

Location: Spain

GMAT 1: 600 Q47 V25 Verified score

WE:Consulting (Consulting)

GMAT 1: 600 Q47 V25 Verified score

Posts: 4

Kudos: 3

[1]

26 Jul 2017, 07:21

1

Kudos

Bookmarks

Bunuel

tonebeeze

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

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

Answer: D.

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 \(2^{2^3}=2^8\)?

Thank you in advance!

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 19 May 2025

Posts: 101,532

Own Kudos:

725,560

[3]

Given Kudos: 93,558

Products:

Expert

Expert reply

Posts: 101,532

Kudos: 725,560

[3]

26 Jul 2017, 07:31

2

Kudos

1

Bookmarks

guillemgc

Bunuel

tonebeeze

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

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

Answer: D.

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 \(2^{2^3}=2^8\)?

Thank you in advance!

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).

JeffTargetTestPrep

Target Test Prep Representative

Joined: 04 Mar 2011

Last visit: 05 Jan 2024

Posts: 3,007

Own Kudos:

7,746

[1]

Given Kudos: 1,646

Status:Head GMAT Instructor

Affiliations: Target Test Prep

Expert

Expert reply

Posts: 3,007

Kudos: 7,746

[1]

09 Aug 2017, 13:27

1

Kudos

Bookmarks

tonebeeze

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.

Answer: D

aaliyahkhalifa

Joined: 10 Oct 2018

Last visit: 11 Feb 2021

Posts: 7

Own Kudos:

2

[1]

Given Kudos: 51

Posts: 7

Kudos: 2

[1]

06 Nov 2018, 00:58

1

Kudos

Bookmarks

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?

EMPOWERgmatRichC

Major Poster

Joined: 19 Dec 2014

Last visit: 31 Dec 2023

Posts: 21,791

Own Kudos:

12,375

Given Kudos: 450

Status:GMAT Assassin/Co-Founder

Affiliations: EMPOWERgmat

Location: United States (CA)

GMAT 1: 800 Q51 V49 Verified score

GRE 1: Q170 V170

Expert

Expert reply

GMAT 1: 800 Q51 V49 Verified score

GRE 1: Q170 V170

Posts: 21,791

Kudos: 12,375

06 Nov 2018, 13:00

Kudos

Bookmarks

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

Gilborg

Joined: 07 Nov 2019

Last visit: 13 May 2021

Posts: 8

Own Kudos:

7

[1]

Given Kudos: 6

Location: Germany

Schools: INSEAD Jan'21 Intake (I) CBS '22 (S)

GMAT 1: 700 Q46 V40 Verified score

GPA: 2.8

Schools: INSEAD Jan'21 Intake (I) CBS '22 (S)

GMAT 1: 700 Q46 V40 Verified score

Posts: 8

Kudos: 7

[1]

12 Feb 2020, 04:56

1

Kudos

Bookmarks

EMPOWERgmatRichC

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

Final Answer:

Show Spoiler

D

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?

EMPOWERgmatRichC

Major Poster

Joined: 19 Dec 2014

Last visit: 31 Dec 2023

Posts: 21,791

Own Kudos:

12,375

Given Kudos: 450

Status:GMAT Assassin/Co-Founder

Affiliations: EMPOWERgmat

Location: United States (CA)

GMAT 1: 800 Q51 V49 Verified score

GRE 1: Q170 V170

Expert

Expert reply

GMAT 1: 800 Q51 V49 Verified score

GRE 1: Q170 V170

Posts: 21,791

Kudos: 12,375

22 Jun 2020, 18:28

Kudos

Bookmarks

Gilborg

EMPOWERgmatRichC

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

Final Answer:

Show Spoiler

D

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?

Hi Gilborg,

You will get the SAME answer (re: the correct one) with all of those examples. When using non-integers, the math gets really 'ugly' though - and you would need an online 'exponent calculator' to prove what the correct answer was. In terms of using X = Y = -1, here's what you'd end up with:

Numerator: 2^(-1 + -1)^2 = 2^(-2)^2 = 2^4 = 16
Denominator: 2^(-1 - -1)^2 = 2^(0)^2 = 2^0 = 1

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

Baan1997

Joined: 07 Feb 2021

Last visit: 10 Mar 2021

Posts: 1

Own Kudos:

2

[2]

Given Kudos: 9

Posts: 1

Kudos: 2

[2]

10 Feb 2021, 04:37

2

Kudos

Bookmarks

Was it just me who thought that it wasn't clear the the (x+y)^2 was a term multiplied by 2 rather than its exponent. I was so damn confused and trying to solve it. It was the only question that I was totally bewildered about it.

CrackverbalGMAT

Major Poster

Joined: 03 Oct 2013

Last visit: 19 May 2025

Posts: 4,849

Own Kudos:

8,507

Given Kudos: 226

Affiliations: CrackVerbal

Location: India

Expert

Expert reply

Posts: 4,849

Kudos: 8,507

22 May 2021, 04:33

Kudos

Bookmarks

Solution:

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

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

= 2^(4xy)

=2^4 =16 (option d)

Devmitra Sen
GMAT SME

Bambi2021

Joined: 13 Mar 2021

Last visit: 23 Dec 2021

Posts: 321

Own Kudos:

128

Given Kudos: 226

Posts: 321

Kudos: 128

27 Jun 2021, 19:03

Kudos

Bookmarks

Baan1997

Was it just me who thought that it wasn't clear the the (x+y)^2 was a term multiplied by 2 rather than its exponent. I was so damn confused and trying to solve it. It was the only question that I was totally bewildered about it.

I made the exact same mistake. Suppose you took the official practice exam #1 and got this question too? It was really unclear that the parentheses were exponents.

AlbertaYEG

Joined: 02 May 2022

Last visit: 05 May 2022

Posts: 2

Given Kudos: 1

Posts: 2

Kudos: 0

02 May 2022, 12:20

Kudos

Bookmarks

Baan1997

Was it just me who thought that it wasn't clear the the (x+y)^2 was a term multiplied by 2 rather than its exponent. I was so damn confused and trying to solve it. It was the only question that I was totally bewildered about it.

I am so glad I found this.....it really confused me. It was not clear to me they were exponents either...

BrushMyQuant

Tutor

Joined: 05 Apr 2011

Last visit: 14 May 2025

Posts: 2,205

Own Kudos:

2,418

Given Kudos: 100

Status:Tutor - BrushMyQuant

Location: India

Concentration: Finance, Marketing

Schools: XLRI (A)

GMAT 1: 700 Q51 V31 Verified score

GPA: 3

WE:Information Technology (Computer Software)

Expert

Expert reply

Schools: XLRI (A)

GMAT 1: 700 Q51 V31 Verified score

Posts: 2,205

Kudos: 2,418

15 Jan 2023, 04:11

Kudos

Bookmarks

Given that xy = 1, we need to find the value of \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\)

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

Using \(a^2 - b^2 = (a+b)*(a-b)\) we get
=> \(2^{(x + y + x - y) * (x + y -( x - y))\)
=> \(2^{2x * 2y}\) = \(2^{4xy}\) = \(2^{4*1}\) = 16

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Exponents

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

Prep Toolkit

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

My Rewards