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If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
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08 May 2022, 08:26
08 May 2022, 08:26
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If xy = 1, what is the value of \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\) ?
A. 2
B. 4
C. 8
D. 16
E. 32
A. 2
B. 4
C. 8
D. 16
E. 32
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
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08 May 2022, 11:22
08 May 2022, 11:22
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If xy = 1, then x=y=1
Inputting this info in the eq we get num = 2^(1+1)^2 ----> 2^4 = 16 and Denominator = 2^(1-1)^2 = 2^0 = 1
Hence correct ans is option D
Inputting this info in the eq we get num = 2^(1+1)^2 ----> 2^4 = 16 and Denominator = 2^(1-1)^2 = 2^0 = 1
Hence correct ans is option D
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Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
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15 Jan 2023, 03:09
15 Jan 2023, 03:09
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Given that xy = 1, we need to find the value of \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\)
\(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\)
=> \(2^{(x+y)^2 - (x-y)^2}\)
Using \(a^2 - b^2 = (a+b)*(a-b)\) we get
=> \(2^{(x + y + x - y) * (x + y -( x - y))\)
=> \(2^{2x * 2y}\) = \(2^{4xy}\) = \(2^{4*1}\) = 16
So, Answer will be D
Hope it helps!
Watch the following video to learn the Basics of Exponents
\(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\)
=> \(2^{(x+y)^2 - (x-y)^2}\)
Using \(a^2 - b^2 = (a+b)*(a-b)\) we get
=> \(2^{(x + y + x - y) * (x + y -( x - y))\)
=> \(2^{2x * 2y}\) = \(2^{4xy}\) = \(2^{4*1}\) = 16
So, Answer will be D
Hope it helps!
Watch the following video to learn the Basics of Exponents
