Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
is the base of the first logarithm a natural log or a log base 10? –  mathemagician Jan 11 '13 at 10:21
2  
Do you mean $\log_{\log{\sqrt{9-x^2}}}{ x^2} <0$? –  Ron Gordon Jan 11 '13 at 10:46

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
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 $\log_b(\log_c\sqrt{9-x^2})<0\implies\log_c\sqrt{9-x^2}<1\implies x^2>9-c^2$

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.