Solve $ 1 - \sqrt{1 - 8\cdot(\log_{1/4}{x})^2} < 3\cdot \log_{1/4}x $

Question

My answer: $2^\frac{-1}{\sqrt{2}} < x < 1$

Textbook answer: $2^\frac{-12}{17} < x < 1$

The only difference between my resolution and the Textbook one is that I solved by saying that $$\text{(I) }\sqrt{1 - 8\cdot (\log_{1/4}{x})^2} > 1 -3\cdot \log_{1/4}x $$

Is true when II OR III are true: $$\text{(II) } 1 - 3\cdot \log_{1/4}x > 0 \quad \land \quad 1-8\cdot (\log_{1/4}{x})^2 > (1 - 3\cdot \log_{1/4}x)^2 $$

$$\text{(III) } 1 - 3\cdot\log_{1/4}x < 0 \quad \land \quad 1- 8\cdot(\log_{1/4}{x})^2 > 0 $$

And the Textbook said that (I) is true when:

$$ \text{(IV) }1- 8\cdot(\log_{1/4}{x})^2 > (1 - 3\cdot \log_{1/4}x)^2 \ge 0 $$

Sorry if this is a dumb question, any help is appreciated. I can't really understand if my conditions (II or III) and their condition (IV) are both right or if one of them is wrong.

Answer 1

at first we have $$x>0$$ and $$\frac{1}{8}\geq \frac{(\ln(x))^2}{(2\ln(2))^2}$$ this condition gives us $$e^{\ln(2)/\sqrt{2}}\geq x\geq1$$ or $$e^{-\ln(2)/\sqrt{2}}\le x<1$$ now we can solve the inequality $$1+\frac{3\ln(x)}{2\ln(2)}<\sqrt{\left(1-\frac{2\ln(x)^2}{\ln(2)^2}\right)}$$ and we do case work: I)if $$x\le e^{-\frac{2\ln(2)}{3}}$$ is hold then our inequality is true. II) if $$x> e^{-\frac{2\ln(2)}{3}}$$ is hold then we get after squaring and simplifying $$\frac{17}{4}(\ln(x))^2+3\ln(2)\ln(x)<0$$ from the last inequality we obtain $$e^{-12/17\ln(2)}<x<1$$ with the condition above we get finaly $$e^{-2\ln(2)/2}<x<1$$

Answer 2

Initially we have $$\sqrt { 1-8\cdot (\log _{ 1/4 }{ x } )^{ 2 } } <1-3\cdot \log _{ 1/4 } x\\ \log _{ 1/4 }{ x } =t\\ \\ 1-8t^{ 2 }>0\\ \left( t-\frac { 1 }{ 2\sqrt { 2 } } \right) \left( t+\frac { 1 }{ 2\sqrt { 2 } } \right) <0\\ t\in \left( -\frac { 1 }{ 2\sqrt { 2 } } ,\frac { 1 }{ 2\sqrt { 2 } } \right) \\ \log _{ 1/4 }{ x } \in \left( -\frac { 1 }{ 2\sqrt { 2 } } ,\frac { 1 }{ 2\sqrt { 2 } } \right) \\ -\frac { 1 }{ \sqrt { 2 } } <\log _{ 2 }{ x } <\frac { 1 }{ \sqrt { 2 } } \\ { 2 }^{ -\frac { 1 }{ \sqrt { 2 } } }<x<{ 2 }^{ \frac { 1 }{ \sqrt { 2 } } }\\ \\ $$

On the other hand, by solving an inequality we get $$ 1-8t^{ 2 }<1-6t+9t^{ 2 }\\ 17t^{ 2 }-6t>0\\ t\left( 17t-6 \right) >0\\ t\in \left( -\infty ,0 \right) \cup \left( \frac { 6 }{ 17 } ,+\infty \right) \\ \log _{ 1/4 }{ x } \in \left( -\infty ,0 \right) \cup \left( \frac { 6 }{ 17 } ,+\infty \right) \\ -\infty <\log _{ 1/4 }{ x } <0\quad \Rightarrow \quad 0<\log _{ 2 }{ x } <+\infty \quad \Rightarrow \quad 1<x<+\infty \\ \frac { 6 }{ 17 } <\log _{ 1/4 }{ x } <+\infty \quad \Rightarrow \quad -\infty <\log _{ 2 }{ x } <-\frac { 12 }{ 17 } \Rightarrow \quad { 2 }^{ -\frac { 12 }{ 17 } }<x<1\\ \\ x\in \left( { 2 }^{ -\frac { 1 }{ \sqrt { 2 } } },{ 2 }^{ \frac { 1 }{ \sqrt { 2 } } } \right) \cap \left( 1,+\infty \right) \cap \left( { 2 }^{ -\frac { 12 }{ 17 } },1 \right) $$ And note that $${ 2 }^{ -\frac { 1 }{ \sqrt { 2 } } }\approx \quad 0.612547326536\\ { 2 }^{ -\frac { 12 }{ 17 } }\approx 0.61306742163\\ { 2 }^{ \frac { 1 }{ \sqrt { 2 } } }\approx 1.63252691944\\ \\ $$

so the answer is

$$\left( { 2 }^{ -\frac { 12 }{ 17 } },1 \right) $$