$ \log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right| $ one formal way to say we won't choose all solutions in degree 2 equation I have the following :
$$
\log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right| 
$$
After to apply some propierties I have the following:
$$
\log\left|\frac{x^2-4x}{3x-10}\right| = 0
$$
After to apply some operations this equation become in degree 2 equation:
$$
x^2-7x+10 = 0
$$
solution is :
$$
x=2 ;x=5
$$
until here, all right.
Problem is: if we substitute $x$ with number "$5$" is good, but if we substitute with "2" is error. We know that error is obvious, its because with number $2$, logarithm is out of the domain.
But if we get $2$ solutions in this degree $2$ equation, what happened with $x=2$. Did it just disappeared?
And are there some 'formal' way to say why we won't choose number 2, that just say : "we won't choose number 2 because it doesn't make sense "
-------------------------+-------+---------+-----+------
edit :
Here is another exercise to explain that there are some solutions that doesn't work even with absolute values in logarithms arguments. I have the same problem in this another exercise :
$
\log(\sqrt{x+14})+\log(\sqrt{x+7})-\log(1.2)=1 \\ 
$
after some operations I get this 2 grade equation:
$
x^2+21x-46=0
$
Solutions are $x_1=-23$ and $x_2=2$
but again : what happened with $-23$ ?. . does it just disappear ?
note : I'm working in real plane.
Edit:
I have the same problem here :
$
\log(\sqrt{x+14})+\log(\sqrt{x+7})-\log(1.2)=1 \\ 
$
after some operations I get this 2 grade equation:
$
x^2+21x-46=0
$
Solutions are $x_1=-23$ and $x_2=2$
but again : what happened with $-23$ ?. . does it just disappear ?
note : I'm working in real plane.
 A: No, $x=2$ is not out of domain. If $x=2$, you get
$$\ln 2-\ln 4=\ln \left(\frac 24\right)=\ln  \left(\frac 12\right)$$ which is correct.
A: You missed solutions. There are altogether four real solutions: $$2,5,\frac{1\pm\sqrt{41}}{2}$$ To find them, the easiest way is to consider the four intervals for $x$ respectively: $(-\infty,0),(0,\frac{10}{3}),(\frac{10}{3},4),(4,+\infty)$.
A: Cant comment but I believe there's a special criterion you have to satisfy first and foremost when dealing with Logarithmic inequalities.
For example, in $\log|x-4|$ you have to ensure the number being logged (I call it the argument) is always a non-negative number (that is $|x-4|$ in this case).
As a generalisation in $\log_ab$,
$b>0  ,a>0$ and $a$ not equal to $0$ are the conditions your logarithmic inequality must satisfy before you do the actual cancelling of the logs in both the sides.
In our case in $\log|x-4|$, $b>0$ as it is a modulus, but don't forget from  $b>0$ you can get an equation $|x-4|>0$
I am leaving the rest for you, don't forget to do an intersection of the sets
Note: I haven't checked if the numbers are "good" or "bad", lone student seems to be saying they are good enough, but I think what I showed is the "formal proof" to reject a solution if it does not satisfy the original condition we had (i am pretty sure there's another formal name for this kind of rejection) you need to keep this in mind when you are doing inequalities
Note 2: i realised I misread the question but I wrote a lot anyway so I am going to keep  it. as for  the question
$\log\left|\frac{x^2-4x}{3x-10}\right| = 0$
you will get $log 1=0$ and then proceed to cancel the log, however, you are still left with the modulus, so you should be getting 4 answers
