# How to find the domain of long log functions?

I'm looking to know how to find the domain of log functions, combined with other things. I know that when considering domains, we can't allow $1/0$ (undefined), or the square root of a negative number. But I can't seem to work out where to start for this example.

$$f(x) = \frac{1}{2-x} + \ln(3+x(2-x))$$ I think so far I know that $x$ can never be $2$ because $1/(2-2)$ would be undefined. How do I do the next bit: $\ln(3+x(2-x))$?

• Hint: For real arguments and function values the argument of $\ln$ must be $>0.$ In your case you have to solve a quadratic equation. Aug 24 '16 at 12:24

Logarithmic functions do not admit negative arguments (if you are working in the real domain), so what you should do is to see for which values of x $3+x(2-x)>0$ so that you can find where the logarithm is defined.
The general domain for a logarithmic function $\ln{x}$ is $(0,\infty)$.
• The argument of your log function is positive in the $(-1,3)$ domain, and as your first argument of the complete function is defined in all the reals except in the $x=2$ point, then the domain of your function is $(-1,2)\cup(2,3)$. Aug 24 '16 at 12:41
Yes, $\frac{1}{2- x}$ has domain "all numbers except 2". ln(x) has domain "all positive x", all x> 0, so "ln(3+ x(2- x))" has domain 3+ x(2- x)> 0. That can be written as $x(3- x)> 3x- x^2> -3$ or $x^2- 3x< 3$ so $x^2- 3x- 3< 0$. The equation $x^2- 3x- 3= 0$ has roots, by the quadratic formula, $\frac{3\pm\sqrt{9+ 12}}{2}= \frac{3\pm\sqrt{21}}{2}$. Since for x= 0, this is $0^2- 3(0)- 3= -3> 0$. $x^2- 3x- 3< 0$ for all x $\frac{3- \sqrt{21}}{2}< x< \frac{3+ \sqrt{21}}{2}$. That includes x= 0 so we have to remove that and the domain of the original function is $\frac{3- \sqrt{21}}{2}< x< 0$ or $0< x< \frac{3+ \sqrt{21}}{2}$.