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Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?
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27 Dec 2015, 19:43
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Anonamy wrote:
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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Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?
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08 Feb 2017, 19:41
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.
Re: If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?
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24 Dec 2018, 10:28
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71% (00:52) correct 
