$\log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right|$ one formal way to say we won't choose all solutions in degree 2 equation

$$\log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right|$$

After applying some properties we have: $$\log\left|\frac{x^2-4x}{3x-10}\right| = 0.$$

After applying some operations, this equation becomes: $$x^2-7x+10 = 0\\ x=2 ;x=5.$$

The problem is: if we substitute $$x$$ with "$$5$$" it's good, but if we substitute with "$$2$$" there's an obvious error, since $$2$$ is out of the logarithm's domain.

Are there some 'formal' ways 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 the logarithms' arguments:

$$\log(\sqrt{x+14})+\log(\sqrt{x+7})-\log(1.2)=1 \\ \cdots\\ x^2+21x-46=0\\ x_1=-23; x_2=2.$$

But again: what happened with $$-23$$ ? Does it just disappear ?

Note: I'm working in $$\mathbb R.$$

• Why do you say “$2$ is out of the domain”? What part of the original equation does not make sense if you plug in $2$ for $x$? Commented Mar 13, 2021 at 5:12
• $x=\frac{1\pm\sqrt{41}}{2}$ are also solutions since $\log|A|=0\iff A=\color{red}{\pm} 1$. Commented Mar 13, 2021 at 5:18
• @Arturo Magidin for example the first term $\log (2-4)$
– NIN
Commented Mar 13, 2021 at 8:01
• But you don’t have $\log(2-4)$, you have $\log|2-4|$; that’s the logarithm of the absolute value of $2-4$, which is the logarithm of $2$, not of $-2$. Commented Mar 13, 2021 at 11:24
• @Arturo Magidin , so Is it correct to put always all arguments of logarithms inside of absolute value ?
– NIN
Commented Mar 13, 2021 at 16:12

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.

• Thank you. But doesn't it seems a little contrived ? It's correct meanwhile you apply the logarithm propierty. But if you reemplace $\log (2-4)$ it doesn't works.
– NIN
Commented Mar 13, 2021 at 8:35
• @NIN you have absolute value $$\log |2-4|=\log |-2|=\log 2$$ Commented Mar 13, 2021 at 9:46
• Thanks lone , so Is it correct to put always all arguments of logarithms inside of absolute value ?
– NIN
Commented Mar 13, 2021 at 16:13
• @NIN Note that, these are two different things. $$\ln x$$ and $$\ln (|x|)$$ Commented Mar 14, 2021 at 5:41

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

• Thank you Adil. I wanted make you some questions but you told me you can't comment. But really thanks
– NIN
Commented Mar 13, 2021 at 16:24
• I can comment on my own questions thou ;-) Commented Mar 13, 2021 at 18:42
• @NIN I just nearly got a stroke reading my own answer, so I have updated it Commented Mar 13, 2021 at 18:49

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: $$(-\infty,0),(0,\frac{10}{3}),(\frac{10}{3},4),(4,+\infty)$$.

• Wao thank you Hyenazixiao. But I don't understand it.
– NIN
Commented Mar 13, 2021 at 16:22
• In order to solve the equation, you need to get rid of the absolute values. Therefore you need to find the three values (0,10/3 and 4) that make the three absolute values equal to zero, and obtain the four intervals I provided above. Commented Mar 13, 2021 at 17:36