Question

The solution set of the inequality ${\log }_4(3^x-1){\log }_{\frac{1}{4}}\left(\frac{3^x-1}{16}\right)\le \frac{3}{4}$ is

• A. $(0,2]$
• B. $[8,\infty )$
• C. $(2,3)$
• D. $(2,8)$

Answer 1

Answer: (A), (B)

The correct options are
A $[8,\infty )$
B $(0,2]$

${\log }_4(3^x-1){\log }_{\frac{1}{4}}\left(\frac{3^x-1}{16}\right)\le \frac{3}{4}$
$\Rightarrow -{\log }_4(3^x-1){\log }_4\left(\frac{3^x-1}{16}\right)\le \frac{3}{4}[\because {\log }_{1/a}b=-{\log }_{a}b]$
$\Rightarrow {\log }_4(3^x-1)(2-{\log }_4(3^x-1))\le \frac{3}{4}[\because \log \frac{a}{b},\log a^m=m\log a,{\log }_{a}a=1=\log a-\log b]$
let $t={\log }_4(3^x-1)$
$4t^2-8t-3\ge 0$
$\Rightarrow t\in (-\infty ,-3/2)\cup (1/2,\infty )$
Since, $y={\log }_4(3^x-1)$
Therefore, $x\in (0,2]\cup (8,\infty ]$