When I have problem with $\log_b(x^2+2ax+a^2)\geq\log_bk$ I usually just say that $x^2+2ax+a^2\geq k$ and $x^2+2ax+a^2>0$.
I used in example a $(x+a)^2$ to make the things easy.
So I do the intersection and take the answer, what makes sense to me and usually works.
But let us suppose that $ b>1$.
In my example I supposed that $\log(x+3)+\log(x+2)=\log(x^2+5x+6)$.
So I didn't do that $x>-3$ and $x>-2$.
I just did that $x^2-5x+6>=12$ and $x^2-5x+6>=0$
This gives me $x>1$ or $x<-6$
And $\log(x+3)+\log(x+2)\geq\log 12$, gives $x>1$ what is the intersection among $x>-3$ and $x>-2$ and $x^2-5x+6.$
It means that when I should answer a question like
$\log_b(x^2+2ax+a^2)>\log_b(K)$, turning the $\log_b(x^2+2ax+a^2)$ in $\log_b(x+a)+\log_b(x+a)>\log_b(K)$.
So make the intersection between all condictions and results?
How to know if $\log_b(x^2+2ax+a^2)$ is not $\log_b(-x-a)+\log_b(-x-a)$? In inequality that gives a different result.
Why does this happen and how to solve these kinds of questions in the "right way"?
To put it simply: the map $\color{blue}{\rm stuff} \mapsto \log_b \color{blue}{\rm stuff}$ is only defined for $\color{blue}{\rm stuff} > 0$, and everything has to be well-defined a priori.
the first equation is defined only if $x+3 > 0$ and $x + 2 > 0$, that is, $x > -2$ (the stronger condition).
the second equation is defined only if $x^2+5x + 6 > 0$, that is, if $x < -3$ or $x > -2$.