Question

Solve the following inequality:
${\log }_3\frac{2^x-5}{27}\cdot {\log }_{1/3}(2^x-5)<2$

Answer 1

${\log }_3\left(\frac{2^x-5}{27}\right){\log }_{\left(1/3\right)}(2^x-5)<2$

$[{\log }_3(2^x-5)-3{\log }_{3}3][-{\log }_3(2^x-5)]<2$

$-{\left({\log }_3(2^x-5)\right)}^2+3{\log }^{}(2^x-5)+2>0$

$({\log }_3(2^x-5)-2)({\log }_3(2^x-5)-1)>0$

${\log }_3(2^x-5)ϵ(1,2)$

$(2^x-5)ϵ(3,9)$

$2^xϵ(8,14)$

$xϵ({\log }_{2}8,{\log }_{2}14)$

$xϵ(3,{\log }_{2}14)$

Also, $(2^x-5)>0$

$2^x>5$

$x>{\log }_{2}5$

Taking intersection we get $xϵ(3,{\log }_{2}14)$