WAEC - Further Mathematics (2006 - No. 7)
Given that \(\frac{1}{8^{2y - 3y}} = 2^{y + 2}\).
\(\frac{1}{5}\)
\(\frac{7}{8}\)
1
\(1\frac{1}{5}\)
توضیح
\(\frac{1}{8^{2y - 3y}} = \frac{1}{8^{-y}} = 8^{y}\)
\(8^{y} = 2^{y + 2} \implies (2^{3})^{y} = 2^{y + 2}\)
\(3y = y + 2 \implies 2y = 2\)
\(y = 1\)