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Exponent Problem

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by kyabe » Mon May 25, 2009 3:32 pm
Hey,

This problem will be very simple once you simplify the exponents.

2^(X+y)^2 = 2^(x2 + y2 + 2Xy) and likewise
2^(X-y)^2 = 2^(x2 + y2 - 2Xy)

now if you just plugin Xy = 1 u get

2^(X+y)^2 = 2^(x2 + y2 + 2) --- i
2^(X-y)^2 = 2^(x2 + y2 - 2) --- ii

Now break the exponents

since a^mn can be written as (a^m)(a^n) i and ii can be written as

2^(X+y)^2 = (2^x2)(2^y2)(2^2) and --- iii
2^(X-y)^2 = (2^x2)(2^y2)(2^-2) --- iv

therefore iii and iv gives us

2^(X+y)^2 / 2^(X-y)^2 = (2^2) / (2^-2)

or 2^(X+y)^2 / 2^(X-y)^2 = (2^2)(2^2) = 16

Hope it helps :)
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by exhilaration » Mon May 25, 2009 4:00 pm
Absolutely, thanks for the clear explanation!
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by gmatplayer » Mon May 25, 2009 5:05 pm
There is more simple explanation on this:

We know that

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

and

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


so the equation becomes

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

= 2^(4xy)

= 2^4

=16
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by gmatplayer » Mon May 25, 2009 5:05 pm
There is more simple explanation on this:

We know that

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

and

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


so the equation becomes

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

= 2^(4xy)

= 2^4

=16
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by Ian Stewart » Mon May 25, 2009 7:57 pm
Or simpler still, if, as the answer choices indicate, there is an exact numerical answer here *whenever* xy = 1 (there's no answer choice which says 'cannot be determined'), we must get the right answer when we plug in *any* values for x and y that make xy = 1 true. So we can simply plug in x = 1 and y = 1 to get

[2^(1+1)^2] / 2^(1-1)^2 = 2^4/2^0 = 16

Or, algebraically, you might notice that, using our exponent rules:

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

what we have now in the exponent is a difference of squares:

= 2^[(x + y + x - y)(x + y - x + y)]
= 2^[(2x)(2y)]
= 2^(4xy)
= 2^4
= 16

Of course, the algebraic solutions above are very good as well - I'm just suggesting alternative ways to approach the problem.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by kyabe » Mon May 25, 2009 11:04 pm
Ian Stewart wrote:Or simpler still, if, as the answer choices indicate, there is an exact numerical answer here *whenever* xy = 1 (there's no answer choice which says 'cannot be determined'), we must get the right answer when we plug in *any* values for x and y that make xy = 1 true. So we can simply plug in x = 1 and y = 1 to get

[2^(1+1)^2] / 2^(1-1)^2 = 2^4/2^0 = 16

Or, algebraically, you might notice that, using our exponent rules:

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

what we have now in the exponent is a difference of squares:

= 2^[(x + y + x - y)(x + y - x + y)]
= 2^[(2x)(2y)]
= 2^(4xy)
= 2^4
= 16

Of course, the algebraic solutions above are very good as well - I'm just suggesting alternative ways to approach the problem.
Thank Ian.. Never thought that way...

Cheers
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by gmat740 » Tue May 26, 2009 12:09 pm
Or simpler still, if, as the answer choices indicate, there is an exact numerical answer here *whenever* xy = 1 (there's no answer choice which says 'cannot be determined'), we must get the right answer when we plug in *any* values for x and y that make xy = 1 true. So we can simply plug in x = 1 and y = 1 to get


I plugged in x=2 and y=0.5

sometimes we get wrong answer when we plug in values like x=1, may be we don't arrive at a generalized solution
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