Solving $|1 - \ln(1 - |2x| + x)| = |1 - |3x||$

I am asking you for help (that is, just verify my passages and perhaps to right my wrongs) about this exercise.

$$|1 - \ln(1 - |2x| + x)| = |1 - |3x||$$

Now here is what I have done. First of all I have to analyse the lefthand term, so starting from the inner part:

$$1 - |2x| + x = \begin{cases} 1+3x\ \ \text{for}\ \ x < 0 \\\\ 1-x\ \ \text{for}\ \ x > 0 \end{cases}$$

This being said and it's now the existence for the logarithm's turn:

$$\begin{cases} 1+3x > 0 \ \ \text{for}\ \ x > -1/3 \\\\ 1-x >0 \ \ \text{for}\ \ x < 1\end{cases}$$

Whence eventually, intersecting the solutions:

$$\ln(1 - |2x| + x) = \begin{cases} \ln(1+3x)\ \ \text{for}\ \ -1/3 < x \leq 0 \\\\ \ln(1-x)\ \ \text{for}\ \ 0\leq x < 1 \end{cases}$$

Going on, it's about the external absolute value. I thence split into two:

$$|1 - \ln(1+3x)| = \begin{cases} 1 - \ln(1+3x) \ \ \text{for} x < \frac{e-1}{3} \\\\ -1 + \ln(1+3x)\ \ \text{for}\ \ x>\frac{e-1}{3}\end{cases}$$

Yet here the second possibility NEVER happens for we had $0\leq x < 1$ where as the first one is less restrictive than what we had ($-1/3 < x \leq 0$) so we neglect it.

For the second piece:

$$|1 - \ln(1-x)| = \begin{cases} 1 - \ln(1-x)\ \ \text{for}x > 1-e \\\\ -1 + \ln(1-x)\ \ \text{for} \ \ x < 1-e\end{cases}$$

Again the second one never happens, and the first one implies a new restriction. Eventually:

$$|1 - \ln(1 - |2x|+x)| = \begin{cases} 1 - \ln(1+3x)\ \ \text{for} -1/3 < x \leq 0 \\\\ 1 - \ln(1-x) \ \ \text{for}\ \ 1-e \leq x < 1\end{cases}$$

Is this correct so far??

Clearly the second part is easier (I hope), getting in the end

$$1 - |3x| = \begin{cases} 1+3x\ \ \text{for}\ \ x<0 \\\\ 1-3x \ \ \text{for}\ \ x> 0\end{cases}$$

Which splits the external absolute values into two pieces again:

$$|1 + 3x| = \begin{cases} 1+3x\ \ \text{for}\ \ x>-1/3 \\\\ -1+3x \ \ \text{for}\ \ x< -1/3\end{cases}$$

and

$$|1 - 3x| = \begin{cases} 1-3x\ \ \text{for}\ \ x>-1/3 \\\\ -1+3x \ \ \text{for}\ \ x> 1/3\end{cases}$$

Then I guess it's all to unify, and what remains is:

$$|1 - |3x|| = \begin{cases} 1+3x\ \ \text{for}\ \ -1/3 <x<0 \\\\ 1-3x \ \ \text{for}\ \ 0<x< 1/3 \\\\ -1+3x\ \ \text{for}\ \ x>1/3 \cup x<-1/3\end{cases}$$

To get the final equations, I split the interval of the logarithm as follows:

$$1-e<x<1 = 1-e<x<-1/3 \cup -1/3<x<0 \cup 0<x<1/3 \cup 1/3<x<1$$

So the final equation reduces to the following equations

$$ \begin{cases} 1 - \ln(1+3x) = 1 + 3x\ \ \ \text{for}\ \ \ -1/3 < x < 0 \\\\ 1 - \ln(1-x) = 1 + 3x\ \ \ \text{for}\ \ \ -1/3 < x < 0 \\\\ 1 - \ln(1-x) = 1 - 3x\ \ \ \text{for}\ \ \ 0 < x < 1/3 \\\\ 1 - \ln(1-x) = -1 + 3x\ \ \ \text{for}\ \ \ 1-e < x < -1/3 \cup 1/3 < x < 1 \\\\ \end{cases} $$

FINAL EDIT

P.s. I do know there is a unique solution which is $x = 0$, it was rather intuibile. But I am demanding if, supposing we do not know anything about the solutions, the method I have used is right or not.

Very good, but there were some mistakes which affected your work.

Yet here the second possibility NEVER happens for we had $0\leqslant x<1$ where as the first one is less restrictive than what we had ($−1/3<x\leqslant0$) so we neglect it.

It's true that the second case doesn't happen, although I don't quite understand your reasoning here. Note that you're currently considering the case where $−1/3<x\leqslant0$. Your work for the absolute value of $|1-\ln(1+3x)|$ seems to split further depending on whether $x\leqslant\frac{e-1}{3}$ or $x\geqslant\frac{e-1}{3}$. But since the entire current interval $−1/3<x\leqslant0$ satisfies $x\leqslant\frac{e-1}{3}$, the second subcase indeed doesn't happen, and thus we have that $|1-\ln(1-|2x|+x)|=1-\ln(1+3x)$ for all $−1/3<x\leqslant0$.

Again the second one never happens, and the first one implies a new restriction. Eventually: $$|1-\ln(1-|2x|+x)| = \begin{cases} 1-\ln(1+3x), \text{ for } -1/3<x\leqslant0 \\ 1-\ln(1-x), \text{ for } 1-e\leqslant x<1 \end{cases}$$

Almost right, except for one thing: the interval for the second case is $0\leqslant x<1$, because that was the original interval for this case.

Which splits the external absolute values into two pieces again: $$|1+3x| = \begin{cases} 1+3x, \text{ for } x>-1/3 \\ -1+3x, \text{ for } x<-1/3 \end{cases}$$

In the second case, the expression becomes $-1-3x$.

and $$|1-3x| = \begin{cases} 1-3x, \text{ for } x>-1/3 \\ -1+3x, \text{ for } x>1/3 \end{cases}$$

The inequality in the first case must be $x<1/3$.

Then I guess it's all to unify, and what remains is:

After correcting the two mistakes that I've pointed out, you should have teh following four pieces: $$|1-|3x|| = \begin{cases} -1-3x, \text{ for } x\leqslant-1/3 \\ 1+3x, \text{ for } -1/3\leqslant x\leqslant0 \\ 1-3x, \text{ for } 0\leqslant x\leqslant1/3 \\ -1+3x, \text{ for } 1/3\leqslant x \\ \end{cases}$$ (And I know that including the endpoints in both cases is bad style.)

And before proceeding, we should note that the first case is irrelevant because the domain of the other side of the equation is $-1/3<x<1$. and for the same reason, the last case will be cut off at $1$ too.

So the final equation reduces to the following equations …

What you wrote is a bit self-contradictory, because you have two different equations for the same interval $-1/3<x<0$. And parts that go outside of the domain of the equation shouldn't be there either. So you should end up with three cases only: for $-1/3<x\leqslant0$, for $0\leqslant x\leqslant1/3$, and for $1/3\leqslant x<1$. Set up those equations and see how we can demonstrate that they don't have solutions other than $x=0$.