log(2(-1)-2)^2 actually possible?

**Latsabb** · July 10th 2012, 12:30 PM

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.

**ebaines** · July 10th 2012, 12:52 PM

The solution x = -1 is indeed correct. Not sure why you think the first log is "impossible" for x = -1. Make the substitution for the left hand side and you get:

log(2x-2)^2 = log(2(-1)-2)^2 = log(-4^2) = log 16. Note that log(-4^2) is NOT the same as 2 log(-4).

For the right hand side: 4 log(1-x) = 4 log(2) = log 16.

**Latsabb** · July 10th 2012, 12:58 PM

I thought that since the number in the parenthesis was a negative, it automatically counted it out. Looks like I am quite sketchy on the details of all of this, since the ^2 wasnt in parenthesis.

**Reckoner** · July 10th 2012, 01:04 PM

Originally Posted by Latsabb

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.

The problem is that the property $\log x^n = n\log x$ only works when $x$ is positive. Since squaring a nonzero number always yields a positive number, we don't actually know whether $2x-2$ is positive or negative. You can correct this problem by using absolute value bars:

$\lg(2x-2)^2 = 4\lg(1-x)$

$\Rightarrow2\lg|2x-2| = 4\lg(1-x)$

$\Rightarrow\lg|2x-2| = 2\lg(1-x)$

$\Rightarrow\lg|2x-2| = \lg(1-x)^2$

$\Rightarrow|2x-2| = (1-x)^2$

This gives two equations:

$2x-2 = -(1-x)^2 \Rightarrow x^2-1=0\Rightarrow x=\pm1$

and

$2x-2 = (1-x)^2 \Rightarrow x^2-4x+3=0\Rightarrow x=1, 3$

1 and 3 are extraneous, so -1 is the only solution.

Originally Posted by Latsabb

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.

No, the first log is defined, because its argument (after squaring) is positive. Let's try it:

Let $x = -1.$ The left-hand side becomes

$\lg(2x-2)^2 = \lg(-4)^2 = \lg16$

and the right-hand side becomes

$4\lg(1-x) = 4\lg2 = \lg16$

and the equation is satisfied.

Originally Posted by Latsabb

Am I completely off base, or is the book, and the person that wrote that response wrong?

If you see something in their explanation that looks incorrect, you can point it out. But I don't see anything wrong with it.

**Reckoner** · July 10th 2012, 01:10 PM

Originally Posted by Latsabb

I thought that since the number in the parenthesis was a negative, it automatically counted it out. Looks like I am quite sketchy on the details of all of this, since the ^2 wasnt in parenthesis.

I understand. There's some admittedly unfortunate notation that many texts use (that I have used myself in the post above) where the function parentheses are omitted for logarithms (and usually for trig functions too). So $\log(x)$ is written as $\log x$ and $\log\left(x^2\right)$ is written as $\log x^2.$ When your book writes $\lg(2x-2)^2,$ I would interpret that as saying $\lg\left([2x-2]^2\right)$ instead of $\left[\lg(2x-2)\right]^2.$

**Plato** · July 10th 2012, 01:11 PM

Originally Posted by Latsabb

I thought that since the number in the parenthesis was a negative, it automatically counted it out. Looks like I am quite sketchy on the details of all of this, since the ^2 wasnt in parenthesis.

I would argue that $\log(2x-2)^2$ is improper notation.

Many computer algebra systems see that as $[\log(2x-2)]^2$ which is clearly undefined for $x=-1$ .

It should be written as $\log[(2x-2)^2]$ which is defined for $x=-1$ .

It should be noted that $\log(x^2)$ can be written as $2\log(|x|)$ that preserves domain.

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