Log equation question

1. Jan 14, 2009

chops369

1. The problem statement, all variables and given/known data
Solve for x:
2log3x + log3(x-1) = log32 + log36x

2. Relevant equations
N/A

3. The attempt at a solution
Ok, so I tried to solve it and I came up with this and got stuck.

log3(x3-x2) = log312x
Even if I cancelled out the logs it would be x3-x2 = 12x and then how would I solve?

Last edited: Jan 14, 2009
2. Jan 14, 2009

tiny-tim

Welcome to PF!

Hi chops369! Welcome to PF!
Divide both sides by x.

3. Jan 14, 2009

chops369

Ohhhh, I see now. I didn't know you had to factor. Thanks a lot.

4. Jan 15, 2009

romolo

Careful of dividing by x! It's OK in this problem since x=0 is an extraneous solution, but you do run the risk of losing a solution.

5. Jan 16, 2009

Дьявол

It is better if you factor x:
x3-x2-12x=0

x(x2-x-12)=0

One of the solutions is x1=0

The other two come from the quadratic equation.

As you can see zero, can't be solution because if you put the 0 in the first equation, it would not be possible since log30 doesn't exist.

Regards.