If (2^{2x – 3y} = 32) and (log_{y} x = 2), find the values of x and y.
Explanation
(2^{2x – 3y} = 32 implies 2^{2x – 3y} = 2^{5})
(implies 2x – 3y = 5 …., (1))
(log_{y} x = 2 implies x = y^{2} …. (2))
Putting (2) into (1), we have
(2y^{2} – 3y = 5 implies 2y^{2} – 3y – 5 = 0)
(2y^{2} – 5y + 2y – 5 = 0)
(y(2y – 5) + 1(2y – 5) = 0)
((2y – 5)(y + 1) = 0)
(implies y = frac{5}{2} ; y = -1)
(therefore x = frac{25}{4}; x = 1)