Question

Solve the following inequality:
${\log }_2(4^x-{5.2}^x+2)>2$

Answer 1

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

Here, $1$

$4^x-5.2^x+2>0$

$2^xϵ(-\infty ,\frac{5-\sqrt{17}}{2})\bigcup (\frac{5+\sqrt{17}}{2},\infty )$

$2^xϵ(0,\frac{5-\sqrt{17}}{2})\bigcup (\frac{5+\sqrt{17}}{2},\infty )$

$xϵ(-\infty ,{\log }_2\left(\frac{5-\sqrt{17}}{2}\right))\bigcup ({\log }_2\left(\frac{5-\sqrt{17}}{2}\right),\infty )$

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

$4^x-5.2^x+2>4$

$4^x-5.2^x-2>0$

$2^xϵ(-\infty ,\frac{5-\sqrt{33}}{2})\bigcup (\frac{5+\sqrt{33}}{2},\infty )$

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

Taking intersection we get $xϵ({\log }_2\left(\frac{5+\sqrt{33}}{2}\right),\infty )$