# Analytically find the domain of a logarithmic function?

I'm taking pre-calc and I'm already falling behind this semester.

I'm hoping someone could give me a simple explanation on how to solve these types of problems:

$$f(x) = \log_5(4-x^2)$$

I have the answer, but I don't know how to get to it exactly.

I think I factor whatever is inside of the log. But then what's the point of the log?

Here's a few more problems that are similar:

$$f(x) = \log(x^2 - 13x + 36)$$

$$f(x) = \ln|7 + 28x|$$

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Logarithm is defined just on POSITIVE reals.. – uforoboa Sep 26 '12 at 0:45
this is just a clever ruse to make you do an inequality of some expression of x and then see what inequality that implies about x – binn Sep 26 '12 at 0:46
You wrote "these types of problems", followed by $f(x)=\log_5(4-x^2)$. But "$f(x)=\log_5(4-x^2)$" doesn't state any problem. If there's something above that that says "Simplify the following", then you'd have stated a math problem. If there's something above that that says "Find the domains of the following functions", then you'd have stated a DIFFERENT math problem. And if there's something above that says something else, you'd have yet another math problem. Before anyone can be sure what problem you're asking about, you need to give us that additional information. – Michael Hardy Sep 26 '12 at 1:54
Sorry about that Michael. I thought it was clear enough that the title explained what the problem was: "Analytically find the domain of a logarithmic function?". That's all the instructions on the paper. – An Alien Sep 26 '12 at 2:03

If $f : I \subset \mathbb{R} \to \mathbb{R}$ is a function that maps an interval of the real numbers to the real numbers and if you want to know the domain of $f$ and there's no restriction you assume that the domain is the maximum set where the function can be defined. In other words, the interval $I$ will be the maximum set where $f$ can be defined.

For instance, what means the function $\log_5$? This function, when applied to a number $x$ gives the number $y$ such that $5^y = x$. But $5$ raised to any power will always give a strictly positive number then the maximum set where it makes sense of talking about $\log_5$ is the set of positive real numbers.

However, you are composing $\log_5$ with $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 4-x^2$. The domain of the composition will be then the subset of the domain of $f$ that the function $f$ maps to the domain of the function $\log_5$, in other words, you want all real $x$ such that:

$4 - x^2 > 0$

Of course this implies $x^2 < 4$ and so $-2<x<2$. Then the domain of the function $\log_5(4-x^2)$ is the set $I = \{x\in \mathbb{R} \mid -2<x<2\}$.

Don't preocupate if you didn't understand what I meant by the domain of the composition and so on, try to understand this case and then study those topics deeper. I think this way you'll be fine.

I hope this answer helps you somehow. Good luck.

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When we work with $\ln(f)$ in which $f=f(x)$ then we should care about this point that $f(x)>0$. Here, you get a function $$f(x)=\ln|7+28x|$$ Since there is an absolute value in $\ln(...)$ so, the only job we can do is to make $7+28x$ non-zero. This means that $x\mathbb R, ~~x\neq\frac{-7}{28}=\frac{-1}4$.

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Nicely argued! +1 – amWhy Nov 18 '13 at 14:33