First of all, I am sorry if this is a question too simple or stupid.
Consider the equation:
$$ \log((x+2)^2) = 2 \log(5) $$
If I apply the logarithm law $ \log_a(b^c) = c \log_a(b) $
$$ \begin{align} 2 \log(x+2) & = 2 \log(5) \\ \log(x+2) &= \log(5) \\ x+2 &= 5 \\ x &= 3 \end{align} $$
But I can see that I am missing a solution, $x = -7$. I noticed that
$$ \begin{align} \log((x+2)^2) &= 2 \log(5) \\ \Updownarrow \\ 2 \log(x+2) &= 2 \log(5) \end{align} $$
Is NOT true. The domain of the first equation is $x \in \mathbb{R}$ but the second equation's is $x \geq -2$.
I know the correct solution.
So I understand that this is not an equivalent transformation of the equation. What I don't know is how I should avoid this. Is there something to keep in mind that would help me evade this mistake? Naturally, I wouldn't have noticed the missing solution, unless I checked the domain of the second equation, which I wouldn't really have had a reason for...
The formula $$\log_a b^c = c \log_a b$$ is true only if $b > 0$ (if we assume that $\log_a$ is a real-valued function). Therefore, an alternative method of solution can proceed as follows: $$\log (x+2)^2 = 2 \log 5 = \log 5^2 = \log 25,$$ and because now all the arguments to $\log$ on both sides must be positive, we have $$(x+2)^2 = 25$$ or $$(x+2-5)(x+2+5) = (x-3)(x+7) = 0,$$ and both solutions are found.