How to solve for x in logarithm: log(3x-2) + log(x-1) = log2x

Use the log identity log (a * b) = log a + log b (as long as they have the same base).

Then we can replace the left hand side with log (3x - 2)(x-1) because the bases are the same.

Then the equation becomes log (3x-2)(x-1) = log 2x.

Since both sides have the same base log, we can "remove" them to get (3x-2)(x-1) = 2x.

Then we have a quadratic. Solve:

(3x-2)(x-1) = 2x.
3x^2 - 5x + 2 = 2x.
3x^2 - 7x + 2 = 0.

Factoring, we have (x-2)(3x-1) = 0.
Setting each one to zero, we have x = 2 or x = 1/3. However, x =2 is the only answer because x = 1/3 makes the argument of log (x-1) negative, which can't happen.

(3x-2)(x-1) = 2x
3x^2-7x+2 = 0
(3x-1)(x-2) = 0
x = 1/3 (extraneous) or 2
x = 2

Ok I'm not sure but I think you do this:

3x-2+x-1=2x
4x-3=2x
4x-2x=3
2x=3
x=1.5