1. ## Domain of Logarithm

ok, so i came accross this problem

(don't know how to subscript here...4 is the base)

g(x) = log4((x+3)/x)

and i have to find the domain. now, i know that (x+3)/x has to be greater than 0, and obviously, x cannot equal 0. i also know that the domain is negative infinity to -3 and 0 to positive infinity because at x<-3, we would have a negative divided by a negative which would yield a positive number and at -3<x<=0 it would be a negative number (or 0 which we don't want). i can see all this. it is obvious. but my problem is working this out on paper.

i know that on the denominator x > 0, but when i work out the top, x+3>0 goes to x>-3 is what i get. so that says that x>-3 and x>0...

this doesn't show that the domain is (-infinity, -3) U (0, infinity)...so how do i work that out on paper. I know this will kill me on more complex equations if i don't figure this out.

I would sincerely appreciate any help figuring this out . I would also like to apologize if I posted incorrectly in anyway (this is my first post).

2. Note that

$\displaystyle \frac{x + 3}{x} = 1 + \frac{3}{x}$.

For $\displaystyle \log_4{\left(\frac{x + 3}{x}\right)} = \log_4{\left(1 + \frac{3}{x}\right)}$ to be defined, then $\displaystyle 1 + \frac{3}{x} > 0$ and $\displaystyle x \neq 0$.

Therefore $\displaystyle \frac{3}{x} < 1$.

You need to consider two cases, one where $\displaystyle x > 0$ and one where $\displaystyle x < 0$, since multiplying by a negative number changes the direction of the inequality symbol...

Case 1: $\displaystyle x < 0$.

Then $\displaystyle \frac{3}{x} < 1$

$\displaystyle 3 > x$.

So when $\displaystyle x < 0, x < 3$. Putting this together gives $\displaystyle x < 0$.

Case 2: $\displaystyle x > 0$.

Then $\displaystyle \frac{3}{x} <1$

$\displaystyle 3 < x$.

So when $\displaystyle x > 0, x > 3$. Therefore $\displaystyle x > 3$.

Thus the domain is $\displaystyle x \in (-\infty, 0) \cup (3, \infty)$.

3. Maybe first taking this:

$\displaystyle \frac{x+3}{x} = 0$

You get the critical points as -3 and 0.

$\displaystyle \begin{array}{|c|c|c|c|} \hline &-3<x & -3<x<0 & x > 0 \\ \hline x+3 & -ve & +ve & +ve \\ x & -ve & -ve & +ve \\ \hline \frac{x+3}{x} & +ve & -ve & +ve \\ \hline \end{array}$

So,

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

for x < -3 and x > 0

I'm not so sure about that though...

4. Originally Posted by Unknown008
Maybe first taking this:

$\displaystyle \frac{x+3}{x} = 0$

You get the critical points as -3 and 0.

$\displaystyle \begin{array}{|c|c|c|c|} \hline &-3<x & -3<x<0 & x > 0 \\ \hline x+3 & -ve & +ve & +ve \\ x & -ve & -ve & +ve \\ \hline \frac{x+3}{x} & +ve & -ve & +ve \\ \hline \end{array}$

So,

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

for x < -3 and x > 0

I'm not so sure about that though...
sorry it took me so long to thank you guys...the help is much appreciated. ive been hella busy w/homework so, besides the day i saw the replies, hadnt been back till today.

5. Lets give some meaning to the whole process.

Domain of the function $\displaystyle g(x)=\log_4\left(\frac{x+3}{x}\right)$ is clearly dictated by the logarithm and the fraction beneath it. In order to calculate the function value one must be able to calculate the value of the logarithm. That can be done a) if $\displaystyle x\neq 0$ and b) if and only if the argument of the logarithm belongs to the domain of the logarithm. That gives us the inequality:
$\displaystyle \frac{x+3}{x}>0$
One way to think of this inequality is as follows:
the quotient of the expressions (x+3) and x must be positive and that will be the case only if both expressions have the same sign, meaning they both simultaneously must be either positive or negative. Hence:
$\displaystyle (x+3> 0\quad and \quad x>0)\quad or \quad (x+3< 0 \quad and \quad x<0)$
$\displaystyle (x> -3\quad and \quad x>0)\quad or \quad (x< -3 \quad and \quad x<0)$
$\displaystyle ( x>0 )\quad or \quad (x< -3)$

Since Mathematics is a language one should translate/substitute every 'and' with an intersection, and every 'or' with a union. Hence,
$\displaystyle \langle0,+\infty \rangle \cup \langle -\infty, -3\rangle$ or reordering that $\displaystyle \langle -\infty, -3\rangle \cup \langle0,+\infty \rangle$