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Anonamy
Would someone please break down the steps to solve this question, step by step?


Hi,
the Q is
If \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2\), what is the value of xy?

(A) -1/4
(B) 1/2
(C) 0
(D) 1/4
(E) 1/2..

\(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2\)

the power can be added if multiplied and can be subtracted if divided..
when you take a power to numerator from denominator, you have to just add - sign..
example 1/2^2=2^(-2)..
so \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2\) will become..
\(2^{(x+y)^2-(x-y)^2} = 2\)..
\(2^{(x^2+y^2+2xy-x^2-y^2+2xy)}=2\)..
so\(2^(4xy)=2^1\).
4xy=1 or xy=1/4..
D
Hope it is clear now..
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NoHalfMeasures
If \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2\), what is the value of xy?

(A) -1/4
(B) 1/2
(C) 0
(D) 1/4
(E) 1/2

\(2^{(x^2 +y^2 + 2xy) - (x^2 +y^2 - 2xy)}= 2^1\)

Or, \(2^{x^2 +y^2 + 2xy - x^2 - y^2 + 2xy}= 2^1\)

Or, \(2^{4xy}= 2^1\)

Or, \(4xy = 1\)

So, \(xy = \frac{1}{4}\)

Thus, answer will be (D) \(\frac{1}{4}\) ...
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On solving
2^4xy = 2^1
4xy = 1
xy = 1/4
D
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Have a query: If base & exponent are the same, then the exponents have to be added? Because I came across this rule, a^n.b^n=(ab)^n and applying this in the subjected problem, gave the wrong answer.
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