I think I got it:
Rewrite all of the terms as base 3:
9^(2x) = (3^2)^(2x) = 3^(4x)
(1/3)^(x+2) = 3^(-x-2)
27 = 3^3
(3^x)^(-2) = 3^(-2x)
Now rewrite the original equation:
(3^(4x)) * (3^(-x-2)) = (3^3) * 3^(-2x)
Now simplify by adding exponents as follows:
3^(3x-2) = 3^(3-2x)
Now, take the log of both sides to yield:
(3x-2)*log3 = (3-2x)*log3
Cancel log 3 from both sides. You are left with:
3x-2 = 3-2x
5x=5
x=1
If you substitute x = 1 back in it should check out.