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Problem:

$$ y=\ln((3x-2)^2) $$

State the domain and the coordinates of the point where the curve crosses the x-axis

At first sight, you say that the domain is $x>\frac23$ because $\ln$ is undefined for negative numbers, so you just rearrange $3x-2>0$.

But the input of $\ln$ is squared, which means there are 2 roots, namely $1$ and $\frac13$.

Contradiction:

By the law of logarithms $$ \ln(x^2)=2\ln(x) $$ Therefore, the function for $y$ can be rewritten as $$ y=2\ln(3x-2) $$ The problem is that half the graph disappears. Now that the input isn't squared, $y$ is undefined for $x\le\frac23$ ($3x-2$ becomes negative) and the entire left half is gone.


So what's the answer? How many roots are there? It seems that math is contradicting itself.

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  • $\begingroup$ Hint: for which values of $x$ do we have $(3x-2)^{2}>0$? $\endgroup$ – Will R Dec 4 '15 at 13:16
  • $\begingroup$ @WillR Well I just read a post here about why the square root always returns the positive root, but theoretically (and not really algebraically) both of the roots I suggested in my post would work. $\endgroup$ – Arc676 Dec 4 '15 at 13:22
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    $\begingroup$ There is a difference between the square root function (e.g. always returning a positive root) and reversing a squared operation from an expression (this gives a $\pm$ solution). $\endgroup$ – Ian Miller Dec 4 '15 at 13:24
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    $\begingroup$ @mathreadler Edited to use more parentheses. $\endgroup$ – Arc676 Dec 4 '15 at 13:28
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    $\begingroup$ Very handwavey but might help: All of these laws work once you know the domains are right, you can't figure out the domain from abusing the law. $\endgroup$ – djechlin Dec 4 '15 at 22:01
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Your law of logarithms only works if the domain makes sense. You wouldn't write that $\ln((-4)^2)=2\ln(-4)$ as $\ln(-4)$ isn't defined. What would be better to write is actually:

$$\ln(x^2)=2\ln(|x|)$$

This would lead you to two solutions as:

$$\ln(3x-2)^2=0$$

$$2\ln(|3x-2|)=0$$

$$\ln(|3x-2|)=0$$

$$|3x-2|=1$$

$$3x-2=\pm1$$

$$3x=1,3$$

$$x=\frac{1}{3},1$$

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  • $\begingroup$ I think you mean $\frac13$ and $1$, but the message comes across regardless. Thanks for a good explanation. $\endgroup$ – Arc676 Dec 4 '15 at 13:20
  • $\begingroup$ @Arc676 Yes. Fixed now. Oops. $\endgroup$ – Ian Miller Dec 4 '15 at 13:22
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The error is in your simplicity, it should be $$\ln x^2 = \ln |x|^2 = 2\ln |x|$$ As the square removes the negatives you cannot suddenly shoehorn them in and think it flies. Beyond that it's just a matter to solve for when $(3x-2)\neq 0$

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  • $\begingroup$ +1 nice short explanation but :"shoehorn them in ... it flies" is rather a mixed metaphor. $\endgroup$ – Ethan Bolker Dec 4 '15 at 21:14
  • $\begingroup$ yes, I shoehorned them in ;) $\endgroup$ – Zelos Malum Dec 5 '15 at 1:41

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