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If x^2=3, what is the value of
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26 Aug 2021, 02:03
26 Aug 2021, 02:03
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Question Stats:
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If \(x^2=3\), what is the value of \(\frac{2^{y^2}}{2^{(x+y)^2}}\)\((4^{xy})\) ?
(A) −1/8
(B) −1/2
(C) 1/8
(D) 1/2
(E) 2
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
(A) −1/8
(B) −1/2
(C) 1/8
(D) 1/2
(E) 2
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
Re: If x^2=3, what is the value of
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26 Aug 2021, 07:49
26 Aug 2021, 07:49
1
\(x^2=3\)
\(\frac{2^{y^2}}{2^{(x+y)^2}}\)\((4^{xy})\)
\(\frac{2^{y^2} * 2^{2xy}}{2^{x^2+2xy+y^2}}\)
\(\frac{1}{2{x^2}}\)
\(\frac{1}{8}\)
Answer C
\(\frac{2^{y^2}}{2^{(x+y)^2}}\)\((4^{xy})\)
\(\frac{2^{y^2} * 2^{2xy}}{2^{x^2+2xy+y^2}}\)
\(\frac{1}{2{x^2}}\)
\(\frac{1}{8}\)
Answer C
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If x^2=3, what is the value of
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08 Sep 2022, 11:18
08 Sep 2022, 11:18
1
Given that \(x^2=3\) and we need to find the value of \(\frac{2^{y^2}}{2^{(x+y)^2}}\)\((4^{xy})\)
\(\frac{2^{y^2}}{2^{(x+y)^2}}\)\((4^{xy})\) = \(2^{y^2} * 2^{-(x+y)^2}* 2^{2xy}\)
= \(2^{y^2} * 2^{-(x^2 + 2xy +y^2)}* 2^{2xy}\) = \(2^{y^2} * 2^{-x^2 - 2xy - y^2)}* 2^{2xy}\)
= \(2^{y^2 -x^2 - 2xy - y^2 - 2xy\) = \(2^{-x^2}\) = \(2^{-3}\) (as \(x^2=3\))
= \(\frac{1}{8}\)
So, Answer will be C
Hope it helps!
\(\frac{2^{y^2}}{2^{(x+y)^2}}\)\((4^{xy})\) = \(2^{y^2} * 2^{-(x+y)^2}* 2^{2xy}\)
= \(2^{y^2} * 2^{-(x^2 + 2xy +y^2)}* 2^{2xy}\) = \(2^{y^2} * 2^{-x^2 - 2xy - y^2)}* 2^{2xy}\)
= \(2^{y^2 -x^2 - 2xy - y^2 - 2xy\) = \(2^{-x^2}\) = \(2^{-3}\) (as \(x^2=3\))
= \(\frac{1}{8}\)
So, Answer will be C
Hope it helps!
