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

Hint $
$\begin{align*} 
f(x) &= \log_3\sqrt{(x+3)}−\log_3(9−x^2) \\
 &=  \frac{1}{2}\log_3(x+3) - \log_3((3-x)(3+x)) \\
 &=\frac{1}{2}\log_3(x+3) - (\log_3(3-x)+\log_3(3+x))
\end{align*}$
