2^(x+y)^2 / 2^(x-y)^2
Expand numerator and denominator
2^(x+y)^2 = 2^[x^2+y^2+2xy] = 2^x^2 * 2^y^2 * 2^2xy
2^(x-y)^2 = 2^[x^2+y^2-2xy] = (2^x^2 * 2^y^2) / 2^2xy
Compare,
2^x^2 * 2^y^2 * 2^2xy / (2^x^2 * 2^y^2) / 2^2xy
2^2xy / 1/ 2^2xy
2^2xy * 2^2xy
2^4xy , [xy=1]
2^4 = 16
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Source: Beat The GMAT — Problem Solving |
Ue this rule:
a ^ m / a ^ n = a ^ m-n
Also we know:
(x+y) ^ 2 = x ^2+y^2+2xy
(x-y) ^ 2 = x ^2+y^2-2xy
So using all the three above 2^(x+y)squared / 2^(x-y)squared simplifies to
=2 ^ (x ^2+y^2+2xy - (x ^2+y^2 - 2xy))
= 2 ^ (x ^2 + y^2 + 2xy - x ^2 - y^2 + 2xy))
=2 ^ (2xy+2xy) (x ^ 2 and - x ^ 2 cancels similarly the y ^ 2 and - y ^2)
= 2 ^ (4xy)
= 2 ^ 4 (since xy=1)
= 16
a ^ m / a ^ n = a ^ m-n
Also we know:
(x+y) ^ 2 = x ^2+y^2+2xy
(x-y) ^ 2 = x ^2+y^2-2xy
So using all the three above 2^(x+y)squared / 2^(x-y)squared simplifies to
=2 ^ (x ^2+y^2+2xy - (x ^2+y^2 - 2xy))
= 2 ^ (x ^2 + y^2 + 2xy - x ^2 - y^2 + 2xy))
=2 ^ (2xy+2xy) (x ^ 2 and - x ^ 2 cancels similarly the y ^ 2 and - y ^2)
= 2 ^ (4xy)
= 2 ^ 4 (since xy=1)
= 16
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You can save a bit of time here. Recall that 2^a/2^b = 2^(a-b). Apply that immediately to the fraction we're given, and we see that it's equal to:
2^[(x+y)^2 - (x-y)^2]
In the exponent we just have a difference of squares, so we can rewrite the exponent:
2^[(x+y+x-y)*(x+y-(x-y))] = 2^(2x*2y) = 2^(4xy)
Since xy=1, this is equal to 2^4 = 16.
2^[(x+y)^2 - (x-y)^2]
In the exponent we just have a difference of squares, so we can rewrite the exponent:
2^[(x+y+x-y)*(x+y-(x-y))] = 2^(2x*2y) = 2^(4xy)
Since xy=1, this is equal to 2^4 = 16.
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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