Question

$lo{g}_2(4^x-{5.2}^x+2)>2$.

Answer 1

${\log }_2(4^x-{5.2}^x+2)>2$

$=4^x-{5.2}^x+2>2^2$

$=(2^x{)}^2-{5.2}^x+2>4$

Let $2^x=t$

$t^2-5t-2>0$

$t=\frac{5\pm \sqrt{25+8}}{2}=\frac{5\pm \sqrt{33}}{2}$

$=(t-\left(\frac{5+\sqrt{33}}{2}\right))(t-\left(\frac{5-\sqrt{33}}{2}\right))>0$

(Refer to Image)

$\therefore$ $2^xϵ\underset{negative}{(-\infty ,\frac{51-\sqrt{33}}{2})}t(\frac{5+\sqrt{33}}{2},\infty )$

$\log$ is not defined

$\therefore$ $xϵ({\log }_2\frac{5+\sqrt{33}}{2},{\log }_2\infty )$ .