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

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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

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

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

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?
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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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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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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,
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Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]
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 $$2^{2^3}=2^8$$?

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

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

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Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]
Baan1997 wrote:
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.
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Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]
Baan1997 wrote:
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...
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Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]
Top Contributor
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

Hope it helps!

Watch the following video to learn the Basics of Exponents

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