1. ## Restrictions for logarithms

I missed a lesson on the product and quotient laws for logarithms and am having a little trouble figuring out the restrictions.

For example:

Log (x^2+7x+12) - Log (x^2-9)

= Log (x+3)(x+4) - Log (x-3)(x+3)

= Log [(x+3)(x+4)]/(x-3)(x+3)]

= Log (x+4)/(x-3)

How would I go about finding the restrictions on this logarithm?

I know in total, there are three possibilities: x>-3, x>-4, x>3

And in the back of the book, it shows that x>3 is the only restriction.

What I tried was to plug in each restriction into each separate log to see if it worked, but when I tried it that way I found that both x>3 and x>-4 worked, and that conflicts with the answer in the back of the book.

All help greatly appreciated!

2. The domain of a log is $(0,\infty)$.

3. Originally Posted by youngb11
= Log (x+4)/(x-3)
Firstly $x\neq 3$, this should be clear.

Taking $x<3$ will give you a negative result inside the logarithm, refer to post #2 for the reason this can't be.

4. Originally Posted by pickslides
Firstly $x\neq 3$, this should be clear.
That's what I was thinking, but the back of the book doesn't include that as a restriction.

Originally Posted by pickslides
Taking $x<3$ will give you a negative result inside the logarithm, refer to post #2 for the reason this can't be.
So to find the restrictions, I have to take all the possible restrictions and the final answer will include the restrictions which will not result in a negative inside the logarithm?

5. Whatever x you plug in, you MUST have a value of $f(x)>0$ where f(x) is the expression inside the log.

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

$\displaystyle\log{\left(\frac{(x+3)(x+4)}{(x-3)(x+3)}\right)}\Rightarrow\log{\left(\frac{x+4}{x-3}\right)}$

$x\neq \pm 3, -4$

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

7. Originally Posted by youngb11
That's what I was thinking, but the back of the book doesn't include that as a restriction.
If $x>3$ is the solution

$x= 3 \subset x\leq 3$

8. Once you check the intervals, you will have $(,)\cup (,)$

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