<s>[INST] I am trying to solve the inequality $\log_{\log{\sqrt{9-x^2}}} x^2 <0$ I am trying to solve  the inequality 
$$\log_{\log{\sqrt{9-x^2}}}   x^2 <0.$$
I got $\mathrm{S.S}=(-\sqrt8 ,-1)\cup( 1,\sqrt8)$, but a friend  got  $\mathrm{S.S}=(-1,1)- \{0\}$.
Please, what is true?
 [/INST] We shall consider two cases:
(1) Case ($\log \sqrt {9-x^2} >1$)
$$\log_{\log \sqrt{9-x^2}} x^2 <0 \Rightarrow $$
$$\left \{
\begin{array}{l}
0 \neq x^2 < 1 \\
\log \sqrt {9-x^2} >1 \\
\end{array}
\right. \Leftrightarrow
\left \{
\begin{array}{l}
-1 < x < 1 \quad \mathrm {and} \quad x\neq 0\\
\sqrt {9-x^2} > 10 \\
\end{array}
\right. \Leftrightarrow $$
$$ \Leftrightarrow
\left \{
\begin{array}{l}
-1 < x < 1 \quad \mathrm {and} \quad x\neq 0\\
-x^2 > 91 \\
\end{array}
\right. \Leftrightarrow
\left \{
\begin{array}{l}
-1 < x < 1 \quad \mathrm {and} \quad x\neq 0\\
x^2 < -91 \\
\end{array}
\right. \Leftrightarrow $$
$$ \Leftrightarrow S = \emptyset $$
(2) Case ($0 < \log \sqrt {9-x^2} < 1$)
$$\log_{\log \sqrt{9-x^2}} x^2 <0 \Rightarrow $$
$$\left \{
\begin{array}{l}
x^2 > 1 \\
0 < \log \sqrt {9-x^2} <1 \\
\end{array}
\right. \Leftrightarrow
\left \{
\begin{array}{l}
x< -1 \quad \mathrm{or} \quad x > 1\\
1< \sqrt {9-x^2} < 10 \\
\end{array}
\right. \Leftrightarrow $$
$$ \Leftrightarrow
\left \{
\begin{array}{l}
x< -1 \quad \mathrm {or} \quad x > 1\\
1< 9-x^2 < 100 \\
\end{array}
\right. \Leftrightarrow
\left \{
\begin{array}{l}
x< -1 \quad \mathrm {or} \quad x > 1\\
-8< -x^2 < 91 \\
\end{array}
\right. \Leftrightarrow $$
$$ \Leftrightarrow
\left \{
\begin{array}{l}
x< -1 \quad \mathrm {or} \quad x > 1\\
8>\ x^2 > -91 \\
\end{array}
\right. \Leftrightarrow
\left \{
\begin{array}{l}
x< -1 \quad \mathrm {or} \quad x > 1\\
-\sqrt{8}< x < \sqrt{8} \\
\end{array}
\right. \Leftrightarrow $$
$$ \Leftrightarrow S = (-\sqrt{8},-1) \cup (1,\sqrt{8}) $$
Therefore you are right and not your friend.
</s>