Solve for x over the real numbers:
27-4 3^(x+1)+9^x = 0
Simplify and substitute y = 3^x:
27-4 3^(x+1)+9^x = 27-12×3^x+(3^x)^2 = y^2-12 y+27 = 0:
y^2-12 y+27 = 0
The left hand side factors into a product with two terms:
(y-9) (y-3) = 0
Split into two equations:
y-9 = 0 or y-3 = 0
Add 9 to both sides:
y = 9 or y-3 = 0
Substitute back for y = 3^x:
3^x = 9 or y-3 = 0
9 = 3^2:
3^x = 3^2 or y-3 = 0
Equate exponents of 3 on both sides:
x = 2 or y-3 = 0
Add 3 to both sides:
x = 2 or y = 3
Substitute back for y = 3^x:
x = 2 or 3^x = 3
3 = 3^1:
x = 2 or 3^x = 3^1
Equate exponents of 3 on both sides:
Answer: |x = 2 or x = 1
