Question

Solve the following inequality:
${\log }_4(18-2^x)\cdot {\log }_2\frac{18-2^x}{8}⩽-1$

Answer 1

$\frac{{\log }_2(18-2^x)}{2}({\log }_2(18-2^x)-3)\le -1{\left({\log }_2(18-2^x)\right)}^2-3{\log }_2(18-2^x)+2\le 0({\log }_2(18-2^x)-1)({\log }_2(18-2^x)-2)\le 01\le {\log }_2(18-2^x)\le 22\le (18-2^x)\le 4-16\le -2^x\le -1416\ge 2^x\ge 14{\log }_{2}16\ge x\ge {\log }_{2}14Also,18-2^x>0\Rightarrow {\log }_{2}18>x$

Taking intersection, we get $x\in [{\log }_{2}14,{\log }_{2}18]$