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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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    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.2012-09-26

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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. Then the domain of the function $\log_5(4-x^2)$ is the set $I = \{x\in \mathbb{R} \mid -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! +12013-11-18