Find the domain of 1/sqrt(3x-2) for the expression to produce a real number?

First, we have to notice that the domain for 1/sqrt (3x-2) is the same with the domain for sqrt (3x-2), excepting the values for x which is cancelling the denominator.

Let's find this excepted value for x (we've considered from the beginning that it's just a single value, based on the fact that the expression is a linear equation, with a single solution).

3x-2 = 0

We'll add 2, both sides of the equation:

3x-2+2=2

3x=2

We'll divide by3, both sides:

x = 2/3.

So, the excluded value for x = 2/3.

Now, let's find the domain for sqrt(3x-2).

For sqrt(3x-2) to exists, the expression (3x-2)>0

So, reiterating the same steps to find the excepted value for x, we'll find that x>2/3.

So the domain of definition is the ineterval (2/3, inf).