Solve \(3^{2x} - 3^{x+2} = 3^{x+1} - 27\)
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Correct Answer: Option B
Explanation:
\(3^{2x} - 3^{x+2} = 3^{x+1} - 27\)
= \((3^{x})^{2} - (3^{x}).(3^{2}) = (3^{x}).(3^{1}) - 27\)
Let \(3^{x}\) be B; we have
= \(B^{2} - 9B - 3B + 27 = B^{2} - 12B + 27 = 0\).
Solving the equation, we have B = 3 or 9.
\(3^{x} = 3\) or \(3^{x} = 9\)
\(3^{x} = 3^{1}\) or \(3^{x} = 3^{2}\)
Equating, we have x = 1 or 2.
\(3^{2x} - 3^{x+2} = 3^{x+1} - 27\)
= \((3^{x})^{2} - (3^{x}).(3^{2}) = (3^{x}).(3^{1}) - 27\)
Let \(3^{x}\) be B; we have
= \(B^{2} - 9B - 3B + 27 = B^{2} - 12B + 27 = 0\).
Solving the equation, we have B = 3 or 9.
\(3^{x} = 3\) or \(3^{x} = 9\)
\(3^{x} = 3^{1}\) or \(3^{x} = 3^{2}\)
Equating, we have x = 1 or 2.