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

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 $\mathbb R.$

ryang
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NIN
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    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 $x$? – Arturo Magidin Mar 13 '21 at 05:12
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    $x=\frac{1\pm\sqrt{41}}{2}$ are also solutions since $\log|A|=0\iff A=\color{red}{\pm} 1$. – mathlove Mar 13 '21 at 05:18
  • @Arturo Magidin for example the first term $\log (2-4)$ – NIN Mar 13 '21 at 08:01
  • 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 $-2$. – Arturo Magidin Mar 13 '21 at 11:24
  • @Arturo Magidin , so Is it correct to put always all arguments of logarithms inside of absolute value ? – NIN Mar 13 '21 at 16:12
  • You may or may not. You don’t have to always include them. But here, the statement of the problem does include them. What is not okay is to ignore them if they are there. – Arturo Magidin Mar 13 '21 at 16:16
  • @Arturo thankyou – NIN Mar 13 '21 at 16:18

3 Answers3

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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.

lone student
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  • 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 Mar 13 '21 at 08:35
  • @NIN you have absolute value $$\log |2-4|=\log |-2|=\log 2$$ – lone student Mar 13 '21 at 09:46
  • Thanks lone , so Is it correct to put always all arguments of logarithms inside of absolute value ? – NIN Mar 13 '21 at 16:13
  • @NIN Note that, these are two different things. $$\ln x $$ and $$\ln (|x|)$$ – lone student Mar 14 '21 at 05:41
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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: $(-\infty,0),(0,\frac{10}{3}),(\frac{10}{3},4),(4,+\infty)$.

IntegralLover
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  • Wao thank you Hyenazixiao. But I don't understand it. – NIN Mar 13 '21 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. – IntegralLover Mar 13 '21 at 17:36
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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

Adil Mohammed
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