Maths(F.5)

3^(2x+2) - 10(3^x) + 1 = 0
9(3^2x) - 10(3^x) + 1 = 0
[(3^x) - 1][9(3^x) - 1] = 0
3^x = 1 or 3^x = 1/9
3^x = 3^0 or 3^-2
x = 0 or -2

3^(2x+2)-10(3^x)+1= 0
3^(2x)3^2-10(3^x)+1= 0
3^(2x)[9] – 10(3^x) + 1 = 0
[9] 3^(2x) - 10(3^x) + 1 = 0
[9] [3^(x)] ^2- 10(3^x) + 1 = 0 [note: 3^(2x) = (3^x) (3^x)]

Let k = 3^x
[9] [3^(x)] ^2- 10(3^x) + 1 = 0 becomes 9k^2 – 10k + 1 = 0
This is a quadratic equation; Factorize.
9k^2 – 10k + 1 = 0
(9k – 1) (k – 1) = 0
Either (9k – 1) = 0 or (k – 1) = 0
9k = 1 or k = 1
k = 1/9 or k = 1
Since k = 3^x
3^x = 1/9 or 3^x = 1

Solve for x,
Don’t have to take logarithm, you can see it.
x = -2 (There is minus sign in front of 2)
or x = 0

Check two cases where x = -2 or x =0 if you have time:
Case 1 where x = -2
3^(2x+2)-10(3^x)+1= 0
3^(2(-2)+2)-10(3^(-2))+ 1= 0
3^(-4+2)-10(3^(-2))+ 1= 0
3^(-2)-10(3^(-2))+ 1= 0
1/9 + 10/9 + 1 = 0
1/9 + 10/9 + 9/9 = 0
0 = 0

Case 2 where x = 0:
3^(2x+2)-10(3^x)+1= 0
3^(2(0)+2)-10(3^(0))+ 1= 0
3^(2)-10 (1) + 1= 0
9 – 10 +1= 0
0 = 0

If you don't get 0 = 0, the answer is wrong.