To solve this equation, we can rewrite each term with the same base and then solve for x.

27^(x-2/3) = (3^3)^(x-2/3) = 3^(3(x-2/3)) = 3^(3x-2)

9^(x-1) = (3^2)^(x-1) = 3^(2(x-1)) = 3^(2x-2)

Now our equation becomes:

3^(3x-2) - 3^(2x-2) = 2*3^(3x-1)

We can then combine like terms:

3^(3x-2) - 3^(2x-2) = 23^(3x-1)
3^(3x-2) - 3^(2x-2) - 23^(3x-1) = 0

Now we have a quadratic equation in terms of 3^(x). Let's make a substitution:

Let y = 3^x

The equation becomes:

y^3 - y^2 - 2y = 0

Factorizing this equation, we get:

y(y^2 - y - 2) = 0
y(y - 2)(y + 1) = 0

So, the possible values for y are 0, 2 and -1. However, y cannot be negative.

Therefore, y = 2

Therefore, 3^x = 2

x = log base 3 (2)

x ≈ 0.6309

Therefore, the solution to the equation is x ≈ 0.6309.