Given that \(\frac{1}{8^{2y - 3y}} = 2^{y + 2}\).
\(\frac{1}{8^{2y - 3y}} = \frac{1}{8^{-y}} = 8^{y}\)
\(8^{y} = 2^{y + 2} \implies (2^{3})^{y} = 2^{y + 2}\)
\(3y = y + 2 \implies 2y = 2\)
\(y = 1\)
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