solve. check for extraneous roots.3^2x-4(3)^x+3=0
let y = 3^x , then your equation becomes
y^2 - 4y + 3 = 0
(y-3)(y-1) = 0
y = 3 or y = 1
then
3^x = 3 ---> x = 1
3^x = 1 ---> x = 0
y^2 - 4y + 3 = 0
(y-3)(y-1) = 0
y = 3 or y = 1
then
3^x = 3 ---> x = 1
3^x = 1 ---> x = 0