Solve this equation 3^(2x+2)-3^(x+3)-3^x+3=0

3^(2x+2)-3^(x+3)-3^x+3=0
3².3^2x - 3³.3^x - 3^x + 3 = 0
9.3^2x - 27.3^x - 3^x + 3 = 0
9.3^2x - 28.3^x + 3 = 0
let t = 3^x ( t > 0)
equation
<=> 9t² - 28t + 3 = 0
<=> t = 3 --> 3^x = 3 --> x = 1
<=> t - 1/9 --> 3^x = 1/9 --> 3^x = 3^(-2) --> x = -2

any questions: bmt.kurua@yahoo.com

Ok after a bit of thought because I'm a bit rusty on maths, I have come up with a solution using my method with the laws of indices

3^(2x+2) can also be written as (3^(2x))(3^2). Using further laws, this can be expanded to ((3^x)^2)(3^2)
which is 9(3^x)^2

next, 3^(x+3) becomes (3^x)(3^3) which is 27(3^x)

then we have the -3^x + 3 at the end. Put this all together and it becomes

9(3^x)^2 - 27(3^x) - 3^x + 3 = 0. Now you can see that this is a quadratic equation, with the variable expressed as (3^x). To solve it, substitute 3^x by x.

9x^2 - 27x - x + 3 = 0

=> 9x^2 - 28x + 3 = 0

using the quadratic formula, this becomes

28 +/- SQRT(28^2-(4x9x3)) all over (2 x 9)

28 +/- SQRT(676) all over 18

28 +/- 26 all over 18

which has solutions x = 3 and x = 0.111 recurring ( or 1/9 )

Substitute 3^x back in and we have 3^x = 3 or 3^x = 1/9

3^x = 3 has the solution x = 1.

3^x = 1/9 has the solution x = -2

So the solutions are x = 1 or x = -2.

... 3^(2x+2) - 3^(x+3) - 3^(x) + 3 = 0
or 9 *3^(2x) - 27 * 3^(x) - 3^(x) + 3 = 0
or 9 * [3^x]^2 - 28 * [3^(x)] + 3 = 0
or ( [3^x] - 3 ) ( 9 * [3^x] - 1 ) = 0
or 3^x = 3 → x = 1
or 3^x = 1/9 → x = -2
or x = { 1, - 2 }

let 3^x = u

replace, solve, replace, done.