# 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 $\mathrm{S.S}=(-1,1)- \{0\}$.

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is the base of the first logarithm a natural log or a log base 10? – mathemagician Jan 11 '13 at 10:21
Do you mean $\log_{\log{\sqrt{9-x^2}}}{ x^2} <0$? – Ron Gordon Jan 11 '13 at 10:46

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.

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thank you RicardoCruz – Ahmed Jan 12 '13 at 20:59
@Ahmed. You're welcome:) – RicardoCruz Jan 12 '13 at 21:41

$$\log_{\log_c(\sqrt{9-x^2})}x^2=\frac{\log_b x^2}{\log_b(\log_c\sqrt{9-x^2})}<0$$

Without any loss of generality, we can take base $b>1$

(i)If $\log_b x^2<0 \iff x^2<1$

then we need $\log_b(\log_c\sqrt{9-x^2})>0\implies \log_c\sqrt{9-x^2}>1\implies x^2<9-c^2$

$\implies x^2<min(1,9-c^2)$

Here observe that for real $x, min(1,9-c^2)>x^2>0\implies 9-c^2>0\implies c^2<9$ else there will be no solution.

(ii)If $\log_b x^2>0 \iff x^2>1$

then we need $\log_b(\log_c\sqrt{9-x^2})<0\implies\log_c\sqrt{9-x^2}<1\implies x^2>9-c^2$

$\implies x^2>max(1,9-c^2)$

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