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 $$ f(x) &lt; 0 $$

So I calculated the domain of function which is   $ D: (-3;3)$

However I am still unable to solve $f(x) &lt; 0$

I simplified $  \log_3\sqrt{(x+3)}−\log_3(9−x^2) &lt; 0$    to   $\log_3\frac{\sqrt{(x+3)}}{(9-x^2)} &lt; 0$

which gets me to $$ \frac{\sqrt{(x+3)}}{9-x^2} &lt; 1$$

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
First of all, we need $9-x^2&gt;0\iff-3&lt;x&lt;3$

As $\log_ax$ is increasing function for $a&gt;1$

We need $$\sqrt{x+3}&lt;9-x^2$$

Let $\sqrt{x+3}=y&gt;0\implies x=y^2-3$

$$\implies0&lt;9-(y^2-3)^2-y=-y+6y^2-y^4=-y(1-6y+y^3)$$

As $y&gt;0,$ we need $$y^3-6y+1&lt;0$$
The solution obtained by GeoGebra: