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).