How do i solve for x: (9^2x)/(27^x-2)=81^3x-4

Answer 1

(9^2x)/(27^x-2)=81^3x-4

since 9 = 3^2 and 27 = 3^3, this is

3^(4x) / 3^(3x-6) = 3^(12x-16)
4x-(3x-6) = 12x-16
x+6 = 12x-16
11x = 22
x = 2

check:
9^4/27^0 = 81^2
yep