Solving a logarithmic equation that has an exception to the power rule

The lost step was the firs one: $\log_3({x^2-3})^2=2\log_3(|x^2-3|)$ (always that $|x^2-3|\neq0$. In your particular problem, always that $x\neq \pm\sqrt{3}$).

Now, if $|x^2-3|\neq 0$, then $2=\log_3({x^2-3})^2=2\log_3(|x^2-3|)$ iff $1=\log_3(|x^2-3|)$ iff $3=|x^2-3|$ iff $x=0$ or $x=\pm\sqrt{6}$

So, the problem arises because of this fact: $\text{log}(a^b)=b\text{log}(a)$ only if $a>0$. So, when you brought the 2 down using the power rule, you were assuming that $x^2-3>0$, when now that you see one solution is $x=0$ so it need not be the case that $x^2-3>0$. That is where the zero solution got thrown out. So in this case, changing it to exponential form retains all the solutions.

With more complicated problems, you just have to keep in mind whether or not the rule you want to use will throw out solutions. I hope this helps!

In this case, if you want to preserve the domain ($\mathbb{R}\setminus\{\pm\sqrt3\}$), you have $$ \log_3|x^2-3|=1, $$ which gives the additional solution.