# Challenging Logarithm Question

1. Jun 28, 2012

### Prototype44

1. The problem statement, all variables and given/known data
2log(x-3)^2=log(x-3)+3

2. Relevant equations
x>3

3. The attempt at a solution
2log(x^2-6x+9)-log(x-3)=3
log(x-3)=3
Not sure where to go from here seeing that it factores and becomes x=3 which makes the solution undefined

2. Jun 29, 2012

### Staff: Mentor

You seem to have lost the coefficient of the first log term.
If what you had on the left was log(x2 - 6x + 9) - log(x - 3), then you could simplify this to log[(x2 - 6x + 9)/(x - 3)], but that's not what you have.

Instead of expanding the (x - 3)2 as you did, use the properties of logs to write 2log(x - 3)2 so that it is log(something).

On the right, 3 = log(103) = log(1000).

3. Jun 29, 2012

### eumyang

I'm not sure what the equation exactly is.

First of all, what is the logarithm base? In the Precalculus books that I've seen (I'm in the US), no base means log base 10. However, I think I heard that in other places no base means natural logarithm.

Second, I'm not sure what is squared. Is it just (x-3) that is squared, or should it be like this?
2[log(x-3)]2=log(x-3)+3
(in other words, the entire expression log (x-3) is squared)

If you mean the former, then you lost the coefficient in front of the log(x-3)2 somewhere.

EDIT: Beaten to it.

4. Jun 29, 2012

### HallsofIvy

Using "laws of logarithms", 2log(x-3)^2= 4 log(x- 3) and then
4 log(x- 3)- log(x-3)= 3 log(x-3)= 3 so that log(x-3)= 1.
If the logarthm is base 10 that says x- 3= 10. If it is base e, x- 3= e.

5. Jun 29, 2012

### e^(i Pi)+1=0

That's what I got but it doesn't work. But, treating it as a quadratic, I got $x=10\sqrt{10}+3$, which does work, as well as 13 again. Why is this?

6. Jun 29, 2012

### Staff: Mentor

It would be helpful if you clarified for us exactly what the problem is. A couple of things are not clear to us:
1. Does "log" mean log10 or loge (same as ln). I assumed you meant log10.
2. Is the expression on the left side 2 log( (x - 3)2) or 2 (log(x-3))2? IOW, what is being squared?

7. Jun 29, 2012

### Prototype44

since a base is missing it is assumed to be base 10

it is 2log((x-3))^2 i just transformed it in the question

8. Jul 7, 2012

### Prototype44

I will solve this problem right now satisfying the specifications laid

2log(x-3)^2=log(x-3)+3
log(x-3)^4=log(x-3)+3
log(x-3)^3=3
3log(x-3)=3
log(x-3)=1
x-3=10
x=13

9. Jul 7, 2012

### HallsofIvy

That is exactly what you wrote originally. It does NOT answer Mark44's question and does not make sense: "$f(x)^2$" is meaningless unless you interpret it as $(f(x))^2$. But in your original question, you wrote as log(x^2- 6x+ 9) which is $ln((x-3)^2)$ and could be more easily written as $2ln(x-3)$.

If you really meant $(ln(x-3))^2$, then you have a quadratic equation: let y= ln(x- 3) and the equation becomes $2y^2= y+ 3$. That is, of course, the same as $2y^2- y- 3= (2y- 3)(y+ 1)= 0$ which has roots y= ln(x- 3)= 3/2 and y= ln(x- 3)= -1. From the first, $x-3 = 10^{3/2}$ and $x= 3+ 10^{3/2}$ which is about
34.6. From the second, $x- 3= 10^{-1}= 0.1$ so that x= 3.1.