Algebratically find domain of $y=\ln \frac{3x-1}{x+2}$ I had this question :  $y=\ln \frac{3x-1}{x+2}$ and to find the domain 
I managed to find most of it, that is, I figured out $x\neq -2$ cause of the denominator. I also found $x>1/3$ by setting the expression inside the $\ln()$ to larger than $0$. However, this did not produce the last solution of $x<-2$.
Could someone please guide me on the proper way to algebraically solve for all of the domain?
 A: As already noted we want $$\frac{3x-1}{x+2}>0$$  Now a rational function can only change sign at a zero or vertical asymptote, so we see any sign changes can only occur at the zero of the numerator $x=\frac{1}{3}$ or the zero of the denominator $x=-2$.  Since both of these zeros have multiplicity 1 which is odd, the function will change sign.  We need to look at the intervals $(-\infty, -2)$, $(-2,\frac{1}{3})$, and $(\frac{1}{3}, \infty)$.  If we choose a test value such as $x=0$ we find $$\frac{3\cdot 0-1}{0+2}=\frac{-1}{2}<0$$  Using the observation from above that the function will change sign at $x=\frac{1}{3}$ and $x=-2$ tells us that the domain is $(-\infty, -2)\cup (\frac{1}{3}, \infty)$.  If you are unfamiliar with the concept of multiplicity of a zero, you could test additional points such as $x=-3$ and $x=1$ to determine the sign of the function. 
A: HINT: it must be $$\frac{3x-1}{x+2}>0$$
this is equivalent to $$x>\frac{1}{3}$$ or $$x<-2$$
A: Notice that
$$\frac{3x-1}{x+2}>0$$
When $3x-1,x+2$ are both positive but also when $3x-1,x+2$ are both negative since $$\frac{-1}{-1}=1$$
They are both negative exactly when $x<-2$
A: $\frac{3x-1}{x+2}>0$, which means either x>1/3 or x<-2. The logic term either/or means either you or me, one is guaranteed to become her boyfriend.
