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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 $x$?

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    What does "shouldn't already come out while solving for $\;x\;$" mean, to begin with? You solved a quadratic, substituted and found out only one root fits the original problem's conditions. What else? – 2017-02-02
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    What are you asking, exactly? – 2017-02-02
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    by taking the inverse we are implicitly making an absolute value – 2017-02-02
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    I'm not too sure what you mean. You found there to be two possible solutions, but using knowledge of the ln you are able to determine the solution. Is there a more specific issue you are having? – 2017-02-02
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    It just felt somewhat strange to me. I'm used to dismiss solutions, because they don't make sense in a physical world, like imaginary solutions to a given problem, even if they are mathematical correct. I feel that doing every step rigorously should yield only correct solutions in the realm of pure mathematics. – 2017-02-02
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    All the steps are rigorous. And you correctly wrote that $\ln(a)=\ln(b)\Rightarrow a=b$ but not $\ln(a)=\ln(b)\Leftrightarrow a=b$. This automatically implies that you could get some extra roots and should check all you've got in the end. – 2017-02-02

2 Answers 2

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;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.

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

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    What??${}{}{}{}{}{}{}$ Something must be going on that I've no the slightest idea about... – 2017-02-02
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    @DonAntonio Can you be more concrete in what you are asking? $a=b\Rightarrow a^2=b^2$, but $a=-1,b=1$ is the solution for the second one, not the first. – 2017-02-02
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    @DonAntonio For a trivial example, note that every equation $a \times x = b$ implies $0 \times x = 0$ but the latter equation admits every $x \in \mathbb{R}$ while the former only admits a far smaller range of solutions. – 2017-02-02
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    @Wolfram what I asked the OP, and he didn't answer, is *what is he asking to begin with*. Of course, I can't understand what your answer is addressing... – 2017-02-02
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    The OP is asking why an equation B which he derived from equation A admits solutions which equation A does not admit. – 2017-02-02
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    @TimonG. Thanks, yet I still can't understand how is this related to the question...and, in fact, I don't even understand what the question is asking... – 2017-02-02
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    @TimonG. Really?? Is that *really* what the OP asked? I couldn't have, for the life of me, guessed ... – 2017-02-02
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    @DonAntonio I don't think there is any problem if you don't understand this, because the question is trivial and bad if I got it right. – 2017-02-02
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    Yeah, he got 2 solutions for equation B but only one of those works for equation A. And then he was asking "Shouldn't this (i.e.: the fact that only one of those solutions is permitted) already come out while solving for x?". I think your answer is perfectly fine @Wolfram but I agree that the question was not very clear. – 2017-02-02
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    @Wolfram As you can see in the comments, there are several veteran members of this site that don't really understand what the OP meant to ask. I am one of those...:) – 2017-02-02