If 3^2x-y = 3^x+y = √27, the value of y is = ?

1¹/₂
2¹/₄
3³/₂
41

Answer:- 4

Explanation:-

Solution:

$\eqalign{ & {{\text{3}}^{2x - y}}{\text{ = }}{{\text{3}}^{x + y}}{\text{ = }}\sqrt {{3^3}} = {3^{\frac{3}{2}}} \cr & \Leftrightarrow 2x - y = \frac{3}{2}and \cr & \,\,\,\,\,\,\,\,\,x + y = \frac{3}{2} \cr & \Leftrightarrow 3x = \frac{3}{2} + \frac{3}{2} = 3 \cr & \Leftrightarrow x = 1 \cr & \therefore y = \left( {\frac{3}{2} - 1} \right) = \frac{1}{2} \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

$\eqalign{ & {{\text{3}}^{2x - y}}{\text{ = }}{{\text{3}}^{x + y}}{\text{ = }}\sqrt {{3^3}} = {3^{\frac{3}{2}}} \cr & \Leftrightarrow 2x - y = \frac{3}{2}and \cr & \,\,\,\,\,\,\,\,\,x + y = \frac{3}{2} \cr & \Leftrightarrow 3x = \frac{3}{2} + \frac{3}{2} = 3 \cr & \Leftrightarrow x = 1 \cr & \therefore y = \left( {\frac{3}{2} - 1} \right) = \frac{1}{2} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }