# Math Help - Finding domain of a function with natural log

1. ## Finding domain of a function with natural log

Okay so I am taking calculus 1 and thought I fully understood how to find the domain of functions until I saw one with a natural log on the top.

f(x)= [ln(12-2x)]/[sqrt(x+5)-3]

so far I have deduced that x can not equal -6 because a square root cant be a negative and that x can not equal 4 because the square root of 9 would be 3 and would give us a 0 on the bottom which we cant allow. However the entire top section has me puzzled. Can anyone please explain how to work with the top end of this function.

2. ## Re: Finding domain of a function with natural log

First of all the denominator cannot be equal to zero therefore you have to solve the irrational equation $\sqrt{x+5}-3=0$ and exclude this solution(s). Otherwise $\ln(x)$ is only defined if $x>0$.

3. ## Re: Finding domain of a function with natural log

Originally Posted by Siron
First of all the denominator cannot be equal to zero therefore you have to solve the irrational equation $\sqrt{x+5}-3=0$ and exclude this solution(s). Otherwise $\ln(x)$ is only defined if $x>0$.
Thanks for your response I figured it out already. I just didn't know how to handle the natural log at first. If anyone else is looking at this thread with the same question I had... This is how it is solved.

So for the top portion we know that ln(x)>0 so if we solve for x we get 6 and know that we can not include six because the ln would be equal to 0 when it needs to be greater then. As for the bottom portion I described it in my initial question so the domain would be (in interval notation)

[-5,4)U(4,6)

4. ## Re: Finding domain of a function with natural log

What in the world do you mean by "ln(x)> 0"? There is no reason why the numerator cannot be zero or negative. What is true is that ln(x) is defined only for x> 0. Since you have ln(12- 2x) you must have 12- 2x> 0 which gives x< 6.

5. ## Re: Finding domain of a function with natural log

Yeah sorry about that I meant ln(x), x>0