Logarithm equation cant have x be negative?

I was doing my homework, when I came upon a question which has me a bit puzzled. The equation is quite simple, and is as follows:

log (x + 6) = 2 log x

So, I immediately rewrite this as:

x + 6 = x^2

Subtract x^2 from both sides, and then multiply both sides by -1 to turn it into a quadratic equality equation:

x^2-x-6=0

I run it through the quadratic formula, and get that x=-2 or 3.

However, the book says that the answer should only be 3, not -2. This made me think that maybe a log couldnt have a negative in it at all, which made me scratch my head a bit. I know that you cant flat out have a negative log. However, since it would be 2 log -2, and that would turn into -2^2, I dont see the problem, because the end result would be log 4. Am I completely wrong here? I was never taught that a log could not have a negative at all, but only that the end result could not have a log in the forum of "log - whatever."

Re: Logarithm equation cant have x be negative?

The logarithmic function $\displaystyle y=\log(x)$ is indeed only defined when $\displaystyle x$ is a positive number (and not zero). The logarithmic function is the inverse of the exponential function, that means:

$\displaystyle y=\log(x) \Leftrightarrow 10^y = x$. Now, it's not possible to find a real number $\displaystyle y$ so that $\displaystyle x$ will be negative.

Re: Logarithm equation cant have x be negative?

I think it might have to do with extraneous solutions or using the original equation. For example, if you graph 2*log(x) and log((x)^2), they are are two different graphs. So graphing log(x + 6) = 2*log(x) and log(x + 6) = log((x)^2) yields different solution sets

Re: Logarithm equation cant have x be negative?

$\displaystyle log x^2 = 2 log x$ is only true if x is positive.

Re: Logarithm equation cant have x be negative?

Ok, good to know then. Thank you very much. I have never encountered a situation like this.