Math Help - Log Function Restrictions

1. Log Function Restrictions

Hi there
I need help with the restrictions on the variables of this question:

Simplify. State any restrictions on the variables.

log(x^2+7x+12)/log(x^2-9)

So my answer is: log(x+4/x-3) which is correct.

Now for the restrictions, I have:
x<-4 and x>3

However the back of the book says the answer is only x>3 and not x<-4 for the restrictions of the variables. But when I sub in any number such as -5 into the ORIGINAL log equation, i get a defined and positive log of x value. So? is mine correct or theirs? Explain fully.

2. Originally Posted by skeske1234
Simplify. State any restrictions on the variables.
log(x^2+7x+12)/log(x^2-9)

So my answer is: log(x+4/x-3) which is correct.

Now for the restrictions, I have:
x<-4 and x>3
.
The answer in red is not correct. The logarithm function does not work that way.
[Math]\log \left( {x^2 + 7x + 12} \right) = \log \left( {x + 3} \right) + \log \left( {x + 4} \right)~\&~\log \left( {x^2 - 9} \right) = \log \left( {x + 3} \right) + \log \left( {x - 3} \right)[/tex].

So the way you have written it does not allow the factors to divide out.

But the restriction above in blue is the domain of the problem as you have written it.

$log\left(\frac{x^2+7x+12}{x^2-9}\right)$

or

$\frac{log(x^2+7x+12)}{log(x^2-9)}$

4. Sorry the question is:

log[(x^2+7x+12)/x^2-9)]

5. Originally Posted by skeske1234
Sorry the question is:
log[(x^2+7x+12)/x^2-9)]
In that case the donain is $(-\infty , -4) \cup (3, \infty)$.

6. Originally Posted by skeske1234
Sorry the question is:

log[(x^2+7x+12)/x^2-9)]

$\frac{x+4}{x-3} > 0$