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Can someone help me with this problem from GMATPrep?

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by tonker » Mon Jan 08, 2007 5:17 pm
If xy=1, what is the value of

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

a) 2
b) 4
c) 8
d) 16
e) 32

The answer is d - but can you provide me with the explanation.

Thanks,
Vito
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Re: Can someone help me with this problem from GMATPrep?

by gabriel » Tue Jan 09, 2007 12:19 am
tonker wrote:If xy=1, what is the value of

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

a) 2
b) 4
c) 8
d) 16
e) 32

The answer is d - but can you provide me with the explanation.

Thanks,
Vito

this basically is a problem on exponents the rule that has to be applied is

a^x/ a^y= a ^ ( x-y)

applying it to the given PS
2^(x+y)^2/2^(x-y)^2 =

2^( (x+y)^2-(x-y)^2) = 2 ^ 4xy now, since xy=1
the eqn becomes 2^4=16....hence the answer d
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by BTGmoderatorRO » Sat Sep 16, 2017 1:19 pm
A simple approach can be used to solve this question,
here, xy=1
using the general formula, a^x/ a^y= a ^ ( x-y)
apply this to the given question
2^(x+y)^2/2^(x-y)^2
=2^( (x+y)^2-(x-y)^2) = 2 ^ 4xy now, since xy=1
we have,
2^4=16, which satisfy the option D.
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by ErikaPrepScholar » Tue Sep 19, 2017 10:34 am
Whenever we have two exponents with the same base, we should jump to <b>exponent rules</b>. In this case, we are dividing exponents with the same base, which means we're dealing with the quotient rule:

a^m/a^n = a^(m-n)

We'll go ahead and apply that here:

Image

We can't add exponents with two different bases (like x+y and x-y), so we should expand (x+y)^2 and (x-y)^2:

Image

Simplify:

Image

The correct answer is D.
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by GMATGuruNY » Tue Sep 19, 2017 10:41 am
tonker wrote:If xy=1, what is the value of

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

a) 2
b) 4
c) 8
d) 16
e) 32
Let x=y=1.
Then:
2^(x+y)² / 2^(x-y)² = 2^(1+1)² / 2^(1-1)² = 2�/2� = 16/1 = 16.

The correct answer is D.
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Re: Can someone help me with this problem from GMATPrep?

by Scott@TargetTestPrep » Sun Feb 02, 2020 12:52 pm
tonker wrote: ↑
Mon Jan 08, 2007 5:17 pm
 If xy=1, what is the value of

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

a) 2
b) 4
c) 8
d) 16
e) 32

The answer is d - but can you provide me with the explanation.

Thanks,
Vito
Simplifying the given expression, we have:

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

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

2^(4xy)

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

Answer: D

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