Solve \(2^{(2y + 2)} - 9(2^{y}) = -2\).
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Correct Answer: Option n
Explanation:
\(2^{(2y + 2)} - 9(2^{y}) = -2\)
= \((2^{2})(2^{2y}) - 9(2^{y}) = -2\)
= \(4(2^{y})^{2} - 9(2^{y}) = -2\)
Let \(2^{y} = x\)
\(\implies 4x^{2} - 9x = -2\)
\(4x^{2} - 9x + 2 = 0\)
\(4x^{2} - 8x - x + 2 = 0\)
\(4x(x - 2) - 1(x - 2) = 0\)
\((4x - 1)(x - 2) = 0 \implies 4x = 1; x = \frac{1}{4}\)
or \(x = 2\)
If \(x = 2^{y} = \frac{1}{4} = 2^{-2} \implies y = -2\)
If \(x = 2^{y} = 2^{1} \implies y = 1\)
\(\therefore y = -2 ; y = 1\)
\(2^{(2y + 2)} - 9(2^{y}) = -2\)
= \((2^{2})(2^{2y}) - 9(2^{y}) = -2\)
= \(4(2^{y})^{2} - 9(2^{y}) = -2\)
Let \(2^{y} = x\)
\(\implies 4x^{2} - 9x = -2\)
\(4x^{2} - 9x + 2 = 0\)
\(4x^{2} - 8x - x + 2 = 0\)
\(4x(x - 2) - 1(x - 2) = 0\)
\((4x - 1)(x - 2) = 0 \implies 4x = 1; x = \frac{1}{4}\)
or \(x = 2\)
If \(x = 2^{y} = \frac{1}{4} = 2^{-2} \implies y = -2\)
If \(x = 2^{y} = 2^{1} \implies y = 1\)
\(\therefore y = -2 ; y = 1\)