solve for x : lnx + ln(3x-2)=0

ln x + ln (3x-2) = 0

We will
use the logarithm properties to solve for x.

We know that:
log a + log b = log a*b

Then we will
write:

ln x + ln (3x-2) = ln x*(3x-2) =
0

==> ln (3x^2-2x) =
0

Now we will rewrite into the exponent
form.

==> 3x^2 - 2x = e^0 =
1

==> 3x^2 - 2x -1 =
0

Now we will use the formula to find the
roots.

==> x1= (2 + sqrt(4+4*3) / 2*3 = (2+4)/6 =
1

==> x2= (2-4)/6 = -2/6 = -1/3 ( not defined for ln
x)

Then the only solution is x=
1