Solve \(3^{2x} - 3^{x+2} = 3^{x+1} - 27\)

A.

1 or 0

B.

1 or 2

C.

1 or -2

D.

-1 or 2

Correct answer is B

\(3^{2x} - 3^{x+2} = 3^{x+1} - 27\)

= \((3^{x})^{2} - (3^{x}).(3^{2}) = (3^{x}).(3^{1}) - 27\)

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.