<s>[INST] Find values of $x$, such as $\log_3 \sqrt{x+3}−\log_3(9−x^2) < 0$ The Function is $$f(x) = \log_3\sqrt{(x+3)}−\log_3(9−x^2)$$
and I need to figure out arguments for which $$ f(x) < 0 $$
So I calculated the domain of function which is   $ D: (-3;3)$
However I am still unable to solve $f(x) < 0$
I simplified $  \log_3\sqrt{(x+3)}−\log_3(9−x^2) < 0$    to   $\log_3\frac{\sqrt{(x+3)}}{(9-x^2)} < 0$
which gets me to $$ \frac{\sqrt{(x+3)}}{9-x^2} < 1$$
But the solutions of the above equation are complex numbers and I definitely should get them as my result. So, what I am doing wrong here?
Would apprecite every answer
 [/INST] First of all, we need $9-x^2>0\iff-3<x<3$
As $\log_ax$ is increasing function for $a>1$
We need $$\sqrt{x+3}<9-x^2$$
Let $\sqrt{x+3}=y>0\implies x=y^2-3$
$$\implies0<9-(y^2-3)^2-y=-y+6y^2-y^4=-y(1-6y+y^3)$$
As $y>0,$ we need $$y^3-6y+1<0$$
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