Solve: $3^{2x-2} - 28(3^{x-2}) + 3 = 0$

A.

x = -2 or x = 1

B.

x = 0 or x = -3

C.

x = 2 or x = 1

D.

x = 0 or x = 3

View answer & explanation

Correct answer is D

$3^{2x-2} - 28(3^{x-2}) + 3 = 0$

$\frac{3^{2x}}{3^2} - \frac{28.3^x}{3^2} + 3 = 0$

$\frac{3^{2x}}{9} - \frac{28.3^x}{9} + 3 = 0$

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

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Solve: $3^{2x-2} - 28(3^{x-2}) + 3 = 0$