(3^2x)-(12)*(3^x)+(27)=0?

y = 3^x
y² = 3^(2x)

y² - 12y + 27 = 0
(y-3)(y-9) = 0
y = {3, 9}

3 = 3^x
x = 1

9 = 3^x
x = 2

x = {1, 2}

3 ^ 2x - 12 * 3^x + 27 = 0

(3^x)^2 - 12*(3^x) + 27 = 0

We can now factor it as if it were a quadratic equation,
with A=1, B=12, C=27. Think of (3^x) as your variable.
The solutions are 3 and 9. So

(3^x - 3)(3^x - 9) = 0

This product can only be zero if either of the terms is zero.
So we now have two equations:

1. 3^x - 3 = 0
2. 3^x - 9 = 0

Solving 1:
3^x - 3 = 0
3^x = 3
x * log(3) = log(3)
x = 1

Solving 2:
3^x = 9
x * log(3) = log(9)
x = log(9)/log(3) = 2

So, there are two solutions: x=1 and x=2.

The methods to solve such equations are given in the reference.

3^(2x) - 12 * 3^x + 27 = 0
(3^x)^2 - 12 * 3^x + 27 = 0

Change 3^x = y:
(3^x)^2 - 12 * 3^x + 27 = 0
y^2 - 12y + 27 = 0
y^2 - 3y - 9y + 27 = 0
(y^2 - 3y) - (9y - 27) = 0
y(y - 3) - 9(y - 3) = 0
(y - 3)(y - 9) = 0

y - 3 = 0
y = 3

y - 9 = 0
y = 9

Change y into 3^x:
y = 3
3^x = 3
3^x = 3^1
x = 1

y = 9
3^x = 9
3^x = 3^2
x = 2

∴ x = 1 , 2