$$
\log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right|
$$
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 $\mathbb R.$
Best Answer
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.