You need to verify what are the excluded values for the given function, hence, you need to consider the following condition of existence, such that:

`3x-2 > 0 => 3x > 2 => x > 2/3`

You should evaluate the right hand limit to the function `log(3x-2), ` for `x->2/3, x >2/3` such that:

`lim_(x->2/3, x> 2/3) log(3x - 2) = log 0 = - oo`

Hence, evaluating the right hand limit of the function yields `lim_(x->2/3, x> 2/3) log(3x - 2) = -oo` , hence, the function has a vertical asymptote at `x = 2/3` .

You should verify if there exists a slant asymptote whose equation is y = mx + n such that:

`m = lim_(x->oo) (log(3x - 2))/x = lim_(x->oo) (ln(3x-2))/(x ln 10) =oo/oo`

Using l'Hospital's theorem yields:

`lim_(x->oo) (ln(3x-2))/(x ln 10) = lim_(x->oo) ((ln(3x-2))')/((x ln 10)')`

`m = 1/ln 10*lim_(x->oo) 3/(3x-2) = 1/ln 10*0 = 0`

You should evaluate n such that:

`n = lim_(x->oo) (ln(3x-2))/(ln 10) = oo`

Hence, evaluating the slant asymptote yields that there are not any.

**Hence, evaluating the asymtpotes yields that there exists a vertical asymptote at `x = 2/3` .**

The asymptote of y = log(3x - 2) has to be determined. The asymptote of a curve is a line such that the distance between the line and the curve approaches 0 as the two tend towards infinity.

The logarithm of a number x is defined only for values of x greater than 0. The curve y = log x has a vertical asymptote at x = 0

For y = log(3x - 2) as 3x - 2 approaches 0, a vertical line can be drawn at x = 2/3 such that the distance between x = 2/3 and y = log(3x-2) approaches 0.

The graph of y = log(2x - 3) and its asymptote are provided below: