Exponential equation problem

The problem is: 9^(2x)*27^(3-x)=1/9

I know that 1/9 turns into 9^-1, but what do I do with the 27?

Thanks

9^(2x)*27^(3-x)

= (3^2)^(2x) * (3^3)^(3-x)

= 3^4x * 3^(9-3x)

= 3^(4x + 9 - 3x)

= 3^(x + 9)

= 3^(-2) ; where 1/9 = 1/3^2 = 3^(-2)

x + 9 = -2

x = -11