Which is the right result(s) of this logarithmic equation?

Question

I came across this equation in my textbook and I have 3 methods to solve it, all giving me different answers so I hope you can help me understand which of these is the right one.

$$\ln (x^2) -1 = 0$$

1. My textbook says to factor the equation as: $$[\ln (x) - 1] [\ln (x) + 1] =0$$

But treating it as $a^2-b^2$ just doesn't seems right to me because in my understanding $\ln (x^2)$ it's not equal to $(\ln(x))^2$. Anyways, doing it this way would result in either:

a) $\ln(x)-1=0$ $\qquad$=> $\ln(x) = 1$ $\qquad\quad$=> $\color{red}{x=e}$

b) $\ln(x) + 1 = 0$ $\qquad$=> $\ln(x) = -1$ $\qquad$ => $\color{red}{x=e^{-1}}$

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Adding 1 to both sides of the equation would result in $\quad$ $\ln(x^2)=1\quad$ and from here the other two methods unfold as follows:

2. $\quad$ $x^2=e^1$ $\qquad$ => $\color{red}{x=\pm\sqrt e}$

3. $\quad$$ 2\ln(x)=1/:2\qquad$ => $\quad\ln(x)=\frac{1}{2}$$\qquad$ => $\color{red}{x=e^\frac{1}{2}}$

So what am I doing wrong, which is the right one and why?

Answer 1

Note that by injectivity of log function

$$\ln \left(x^2\right) -1 = 0\iff \ln \left(x^2\right) =1 \iff x^2=e\iff x=\pm\sqrt e$$

Answer 2

First approach confuses: $\ln \left(x^2\right)$ with $(\ln x)^2$.

For the third approach omitted the possibility that $x$ can be negative.

$$\ln \left(x^2\right)=\ln \left(|x|^2\right)=1$$

$$2\ln |x|=1$$

Second approach is correct.