You need to write `4/9` such that:

`4/9 = 2^2/3^2 = (2/3)^2`

Using the negative power property yields:

`4/9 = (2/3)^2 = (3/2)^(-2)`

Raising to 6th power both sides, yields:

`(4/9)^6 = ((3/2)^(-2))^6`

Using the properties of exponents, yields:

`(4/9)^6 = ((3/2))^(-2*6) => (4/9)^6 = ((3/2))^(-12)`

Replacing `((3/2))^(-12)` for `(4/9)^6` in equation, yields:

`(3/2)^(x(2x - 4)) = ((3/2))^(-12)`

Equating the exponents yields:

`x(2x - 4) = -12 => 2x^2 - 4x + 12 = 0`

You need to divide by 2 such that:

`x^2 - 2x + 12 = 0`

Using quadratic formula yields:

`x_(1,2) = (2+-sqrt(4 - 48))/2 => x_(1,2) = (2+-sqrt(-44))/2`

Since `sqrt(-44) !in R => x_(1,2) !in R`

**Hence, evaluating the real solutions to the given exponential equation, yields that there exists no real solutions.**