Roots of ln of a square

Question

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.

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

Answer 1

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:

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

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

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

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

$$3x-2=\pm1$$

$$3x=1,3$$

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

Answer 2

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 $(3x-2)\neq 0$