What do I miss? $\ln(x^2 -4) = \ln(1-4x)$, $x \neq 1$

Question

Solve $$\ln(x^2-4) = \ln(1-4x)$$ =>

1)$$x^2-4 = 1-4x$$ 2) $$x_{1/2} = -5, 1$$ But since $\ln(-3)$ is not defined, only $x=-5$ is a solution. Shouldn't this already come out while solving for $x$?

Answer 1

;TDLR You have to keep in mind that the domain of equation 1) is not all real numbers but only a subset defined by the original equation (its domain).

Note that logarithm is defined as follows

$$\log_{\color{blue}{a}} \color{green}{b}$$ where $\color{blue}{a}, \color{green}{b} > 0$ and $\color{blue}{a} \neq 1$.

Which for your case yields that domain of $x$ for which this equation is solved is given by

$x^2 - 4 > 0$ and $1 - 4x > 0$ Which gives us that

$x \in (-\infty,-2) \cup (2,+\infty)$ and $x \in \left(-\infty,\frac{1}{4}\right)$

Which gives us that the domain is expressed as

$$D = x \in (-\infty,-2)$$

Therefore, only solution $x = -5 \in D$ is in the domain.

Answer 2

No, because if $A$ and $B$ are equalities and $A\Rightarrow B$, $B$ might have some solutions that $A$ doesn't have.