Find values of $x$, such as $\log_3 \sqrt{x+3}−\log_3(9−x^2) < 0$

Question

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

Answer 1

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 $$y^3-6y+1<0$$

Answer 2

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

Answer 3

The solution obtained by GeoGebra:

Answer 4

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,

$$x_2 < x < x_1$$