Question

The solution set of the inequality ${\left({\log }_{9}x\right)}^2\ge {\left[{\log }_3(1-\frac{x}{4})\right]}^2$ contains

• A. $2$
• B. $\frac{4}{5}$
• C. $(0,\frac{4}{5})$
• D. None of the above

Answer 1

Answer: (A), (B), (C)

The correct options are
A $2$
B $\frac{4}{5}$
C $(0,\frac{4}{5})$

${\left({\log }_{9}x\right)}^2\ge {\left[{\log }_3(1-\frac{x}{4})\right]}^2$

$1-\frac{x}{4}>0\Rightarrow x<4$

$x>0$

${\left({\log }_{3^2}x\right)}^2\ge {\left[{\log }_3(1-\frac{x}{4})\right]}^2$

$\Rightarrow {\left[\frac{1}{2}\left({\log }_{3}x\right)\right]}^2\ge {\left[{\log }_3(1-\frac{x}{4})\right]}^2$

$\Rightarrow {\left[\frac{1}{2}\left({\log }_{3}x\right)\right]}^2-{\left[{\log }_3(1-\frac{x}{4})\right]}^2\ge 0$

$\Rightarrow [\left({\log }_3\sqrt{x}\right)-{\log }_3(1-\frac{x}{4})]\times [{\log }_3\sqrt{x}+{\log }_3(1-\frac{x}{4})]\ge 0$

$\Rightarrow \left[{\log }_3\frac{\sqrt{x}}{(1-\frac{x}{4})}\right].\left[{\log }_3\sqrt{x}.(1-\frac{x}{4})\right]\ge 0$

for $x=2\to$

$\left[{\log }_3\frac{\sqrt{x}}{(1-\frac{x}{4})}\right].\left[{\log }_3\sqrt{x}.(1-\frac{x}{4})\right]$

$=\left({\log }_{3}2\sqrt{2}\right).\left({\log }_3\frac{1}{\sqrt{2}}\right)$

$\Rightarrow$ It is less than 0 since ${\log }_3\left(2\sqrt{2}\right)>0$ and ${\log }_3\left(\frac{1}{\sqrt{2}}\right)<0$

For $x=\frac{4}{5}$

$\left({\log }_3\frac{\sqrt{\frac{4}{5}}}{\frac{4}{5}}\right).{\log }_3(\frac{\sqrt{4}}{5}.\frac{4}{5})<0$

Here, $\left({\log }_3\frac{\sqrt{\frac{4}{5}}}{\frac{4}{5}}\right)\to$ Positive Term.

and, ${\log }_3(\frac{\sqrt{4}}{5}.\frac{4}{5})\to$ Negative Term.

$\therefore xϵ(0,\frac{4}{5})$