Hello, yet another question, sorry.

My book has the following equation:

log(2x-2)^{2}=4lg(1-x)

Now, I worked this out, and got that x had to equal 1 or 3. Neither of those are possible, as they give you a negative inside the parenthesis. So, I wrote off the problem, and assumed it was one of the "no solution" problems. Well, I found myself shocked when I looked in the back of the book to check my answers, and it showed that x = -1. Doesnt that make the first log impossible? I cant even get that to work on my calculator.

I assumed that maybe the book made a mistake, so I went online and looked for one of those lists of corrections for books, and found them actually defending the -1 answer, after lots of people had said it was a mistake. Their reasoning was as such:

Many students want to solve this as follows:

lg(2x-2)^{2}=4lg(1-x)

2lg(2x-2)=4lg(1-x)

lg(2x-2)=lg(1-x)^{2}

2x-2=(1-x)^{2}

This equality has the solution as x=1 or x=3.

It is because of this that students lose one of the solutions. With the above, they dont get x=-1, and it is easy to conclude that the quality has no solution, because x=1 and x=3 cannot be used.

The problem is in line number 2 in this solution. The left side is defined when x>1 and the right side is defined when x<1. Meaning no solution.

Am I completely off base, or is the book, and the person that wrote that response wrong? I want to assume they are right, since I am not exactly a math wiz, but I cannot get it to work.

Edit: Sorry, forgot to clip out the Norwegian after I translated it.