Question 1 Report
2. If 2\(^{2x -2y}\) = 32 and log\(_y\) x = 2, find the values of x and y
Leave your answer in this format "+ x,- y"
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
When y = -1
x = (-1)\(^2\)
x = 1
When y = \(\frac{5}{2}\)
x = (\(\frac{5}{2}\))\(^2\) = \(\frac{25}{4}\) = 6\(\frac{1}{4}\).
Answer Details
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