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Find the domain of $f(x)=\ln(3x+2)$

I can find domain, but is it the same for a $\log$ function? And also, do I have to rid the equation of the $\ln$ before I can find the domain? I'm really confused as to how to start out this problem. Please help.

Edit: I got the answer and now I ask a question (irrelevant to my problem but I am curious), why does the real portion of $f(x)=\log x$ resemble the graph of a square root function. In a sense of, starting at a single point and going in two directions: up/down & left/right.

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The $\log$ and $\ln$ functions are pretty much the same thing, because $\ln(x) = \log(x) / \log(e)$, and $\log(e)$ is just some constant. – MJD Aug 21 '12 at 18:48
up vote 9 down vote accepted

Hint: $\log t$ is defined precisely if $t$ is positive. The base is irrelevant.

Why: The explanation depends on how you define the logarithm. Let $b\gt 0$, with $b\ne 1$. Then $\log_b t$ can be defined as the number $w$ such that $b^w=t$. But $b^w$ is always positive, so $t$ must be positive. Thus $\log_b t$ is undefined if $t \le 0$.

To show that $\log_b t$ is defined whenever $t\gt 0$, we need some information about the function $b^w$. Assume that $b\gt 1$. Then $b^w$ is steadily increasing, and is close to $0$ when $w$ is large negative, and is huge when $w$ is large positive. Thus the function $b^w$ takes on every positive value once and only once. A similar argument settles the (much less common) situation where $0\lt b\lt 1$.

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$x\in \mathbb R:x\gt -\frac{2}{3}$? – Austin Broussard Aug 21 '12 at 18:55
@AustinBroussard: Yes, you have got it. I would write $\{x\in\mathbb{R}: x\gt -\frac{2}{3}\}$. – André Nicolas Aug 21 '12 at 19:01
Okay, and this happens because we ignore the imaginary part and get the domain of the real part. See my edit please. – Austin Broussard Aug 21 '12 at 19:01
@AustinBroussard: The familiar $\ln$ or $\log$ of your course is defined only for positive reals, and is real, that is, has imaginary part $0$. So we are not ignoring the imaginary part. When we go to the complex numbers, $\log$ is multiple-valued, so it is no longer a function. As to the resemblance in shape with $\sqrt{x}$, they are both concave downwards. But in the long run $\sqrt{x}$ is enormously larger than $\ln x$. To see this, use your calculator to find $\ln(1000000)$. – André Nicolas Aug 21 '12 at 19:09

What is the domain of $g(t)=\ln(t)$? For which $x$ is $3x+2$ in this domain? There's no need to remove the logarithm first.

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Okay, so if I plot $\log(3x+2)$ (only the real part), I would then find my domain accordingly? – Austin Broussard Aug 21 '12 at 18:53
Well, since you're only considering real numbers, then the domain of any basic logarithm is the positive reals. Your job, then, is to figure out for which $x$ we have $3x+2>0$. – Cameron Buie Aug 21 '12 at 19:19

As this is tagged pre-calculus, I'll assume that you want $y=f(x)$ to be a real number. The first step is: where is $\log x$ undefined? Then, when you answer that question, you can adjust that domain for the transform $x \to 3x+2$.

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$log_a (b)=c$ if and only if $a^c = b$, where $a>0$. Since $a$ is positive, $b$ will also be positive since a positive number to a real exponent will always be positive. Therefore, the expression in the logarithm will always be greater than zero. In your case, this simplifies to $3x+2>0$. Solve for $x$ to get $x>-\frac{2}{3}$. Thus, your domain is $(-2/3, \infty)$.

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