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

\(2y^{2} - 3y = 5 \implies 2y^{2} - 3y - 5 = 0\)

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

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

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

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

\(\therefore x = \frac{25}{4}; x = 1\)

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