Question

Solve the following inequality:
${\log }_4(3^x-1){\log }_{1/4}\frac{3^x-1}{16}⩽\frac{3}{4}$

Answer 1

$-{\log }_4(3^x-1)\left[{\log }_4\right(3^x-1)-2]\le \frac{3}{4}{\log }_4^2(3^x-1)-2\log (3^x-1)+\frac{3}{4}\ge 04{\log }_4^2(3^x-1)-8\log (3^x-1)+3\ge 04{\log }_4^2(3^x-1)-2\log (3^x-1)-6\log (3^x-1)+3\ge 02{\log }_4(3^x-1)\left(2{\log }_4\right(3^x-1)-1)-3\left(2{\log }_4\right(3^x-1)-1)\ge 0\left(2{\log }_4\right(3^x-1)-1)\left(2{\log }_4\right(3^x-1)-3)\ge 0{\log }_4(3^x-1)\in (-\infty ,\frac{1}{2})\cup (\frac{3}{2},\infty )(3^x-1)\in (0,2)\cup (8,\infty )3^x\in (1,3)\cup (9,\infty )x\in (0,1)\cup (2,\infty )$