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

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Math Expert
Joined: 02 Aug 2009
Posts: 6223
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]

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30 May 2016, 03:31
DensetsuNo wrote:
Although I'm pretty much familiar with exponential rules,
and got this question right in the Gmatprep, I've a doubt.

The rules of exponentials tell us that whenever we have $$(n^x)^y$$ the result is n elevated at the x*y,
so why aren't we multiplying (x+y) and (x-y) for 2?

Thanks

Hi if I have been able to understand the doubt...

$$(n^x)^y = n^{xy}$$ ...
$$But n^x*n^y = n^{x+y}.. and ..\frac{n^x}{n^y}= n^{x-y}$$

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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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30 May 2016, 03:57
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

Hi chetan,
sorry I've not been able to explain myself (I'm starting to think that the question has to be even dumber than previously thought :D),

We have
$$\frac {2^{(x + y)^2}} {2^{(x - y)^2}$$
where (x+y) and (x-y) are elevated by 2,
the properties of the exponentials do not tells us that one exponential elevated another exponential equals the product of the two?
eg. $${2^2}^3 = 2^6$$
Manager
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GMAT 1: 730 Q49 V40
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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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30 May 2016, 04:12
Sorry to double post,
I thought a bit through what I was saying and what I was saying is correct when there's one parenthesis,
eg. $$(2^2)^3 = 2^6$$
While if there's no such parenthesis the rules of "calculus" tell us that the power takes priority over the multiplication
and therefore 2^2^3 -> 2^3 = 8 -> 2^8

Doubt cleared!
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Posts: 6
Location: Spain
GMAT 1: 600 Q47 V25
WE: Consulting (Consulting)
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]

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26 Jul 2017, 07:21
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
Joined: 02 Sep 2009
Posts: 47039
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]

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26 Jul 2017, 07:31
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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Joined: 26 Apr 2017
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GMAT 1: 600 Q47 V25
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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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26 Jul 2017, 12:36
Bunuel wrote:
guillemgc wrote:
Bunuel wrote:

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

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).

I understand it now! Thank you so much!

Regards.
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2679
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]

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09 Aug 2017, 13:27
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] 09 Aug 2017, 13:27

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