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

Question

Simplify and write without negative exponents:

3^2(x^-2y^3)^3/4^1/2*x^-4cuberooty^2

Answer 1

(3^x)^2-12*3^x+27=0
let y = 3x , so we have

y^2 - 12y + 27 = 0
(y - 9)(y - 3) = 0
y = 9 or y = 3
then 3^x = 9
3^x = 3^2 -----> x = 2
or
3^x = 3 ----> x = 1