Explanation

2\(^{2x -2y}\) = 32 = 2\(^{2x -2y}\) = 2\(^5\)

2x - 2y = 5 - - - - -  - - -- (1)

 log\(_y\) x = 2,

x = y\(^2\)   - - - - - - -  -(2)

Put x = y\(^2\) into equation (i) 2(y\(^2\)) - 3y = 5

2\(y^2\) - 3y - 5 = 0

2y\(^2\) + 2y - 5y - 5 = 0

2y(y + 1) - 5(y + 1) = 0

(2y - 5)(y + 1) = 0

2y - 5 = 0 or y + 1 = 0

2y = 5 or y = -1

y = \(\frac{5}{2}\) or y = -1

y = \(\frac{5}{2}\) or y = -1

When y = -1

x = (-1)\(^2\)

x = 1

When y =  \(\frac{5}{2}\)

x = (\(\frac{5}{2}\))\(^2\) = \(\frac{25}{4}\) = 6\(\frac{1}{4}\).

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