TITLE: Solving a logarithmic equation that has an exception to the power rule
QUESTION [25 upvotes]: Given the following:
$$\log_3({x^2-3})^2=2$$
If I were to use the power rule, I would do:
$$2\log_3({x^2-3})=2$$
$$\log_3({x^2-3})=1$$
$$3^1=x^2-3$$
$$3+3=x^2$$
$$x=\pm\sqrt6$$
Substituting back into the original equation, it is correct. However, there is a missed answer:
$$x=0$$
Substituting that into the original equation, it is correct as well, but while using the power rule, it has been overlooked.
Obviously, this is a rather simple question, so I could have just done $3^2=(x^2-3)^2$ and solved it like that, in which case the answer of $0$ would appear. However, for more complicated questions, how would one go about doing it? And are there exceptions to this exception?

REPLY [2 votes]: There is NO 'exception to the power rule' here, it's your error in expression handling. The error is similar to saying, if $(x+1)^2 > 9$ then $x+1>3$.
The square in the inner function guarantees the logarithm argument is non-negative, and the equation's domain is all reals with two points excluded: 
$$\{x:(x^2−3)^2 > 0\} = \{x:(x^2−3)\ne 0\} = \mathbb R \setminus \{\pm\sqrt 3\}$$
'Removing' the square cuts-off the domain portion in which $(x^2-3)<0$, leaving
$$\{x:(x^2-3)>0\} = \mathbb R \setminus [-\sqrt 3,\sqrt 3]$$
hence you miss solution(s) inside the interval $(-\sqrt 3,\sqrt 3)$.