Solve: \(3^{2x-2} – 28(3^{x-2}) + 3 = 0\)
- x = -2 or x = 1
- x = 0 or x = -3
- x = 2 or x = 1
-
x = 0 or x = 3
The correct answer is: D
Explanation
\(3^{2x-2} - 28(3^{x-2}) + 3 = 0\)
\(\frac{3^{2x}}{3^2} - \frac{28.3^x}{3^2} + 3 = 0\)
\(\frac{3^{2x}}{9} - \frac{28.3^x}{9} + 3 = 0\)
let p = 3\(^x\)
\(\frac{p^{2}}{9} - \frac{{28p}}{9} + 3 = 0\)
multiply through by 9
p\(^2\) - 28p + 27 = 0
p\(^2\) - p - 27p + 27 = 0
p (p - 1) - 27(p - 1) = 0
(p-1)(p-27) = 0
p = 1 or 27
when p = 1
p = 3\(^x\)
3\(^x\) = 1
3\(^x\) = 3\(^0\)
x = 0
when p = 27
3x = 27
3x = 33
x = 3