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
 A: 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 $$y^3-6y+1<0$$
A: Hint $
$\begin{align*} 
f(x) &= \log_3\sqrt{(x+3)}−\log_3(9−x^2) \\
 &=  \frac{1}{2}\log_3(x+3) - \log_3((3-x)(3+x)) \\
 &=\frac{1}{2}\log_3(x+3) - (\log_3(3-x)+\log_3(3+x))
\end{align*}$
A: Continue with $ \frac{\sqrt{x+3}}{9-x^2} < 1$, yet with the domain restriction  $-3<x<3$. Then, examine the roots of the equation 
$$\sqrt{x+3} = 9-x^2$$
Square and factorize to get
$$(x+3)(x^3-3x^2-9x + 26)=0$$
Since $x>-3$, the valid roots come from the second factor. Let $x= t+1$ to depress the second factor as
$$t^3-12t+15=0$$
which has three real roots. Use the analytical root expressions below for the cubic equation of the form $t^2+pt+q=0$,
$$t_k = 2\sqrt{\frac{-p}3}\cos\left( \frac13\cos^{-1}\left(\frac{3q}{2p}\sqrt{\frac{-3}p} \right) -\frac{2\pi k}3 \right), \>\>\>k=0,1,2$$
The valid solutions for $x$ are
$$x_1=1+4\cos\left(\frac13\cos^{-1}\frac{15}{16}+\frac{\pi}3\right)=2.577$$
$$ x_2=1-4\cos\left(\frac13\cos^{-1}\frac{15}{16}\right)= -2.972$$
Then, it is straightforward to verify that $ \frac{\sqrt{x+3}}{9-x^2} < 1$ if the values of $x$ are in the range,
$$x_2 < x < x_1$$
A: The solution obtained by GeoGebra:

