Question

Solve the following inequality:
$\frac{{\log }_2({3.2}^{x-1}-1)}{x}⩾1$

Answer 1

$letx>0$

$\Rightarrow {\log }_2({3.2}^{x-1}-1)\ge x$

$\Rightarrow {3.2}^{x-1}-1\ge 2^x$

$\Rightarrow \frac{1}{2}{2}^x\ge 1$

$\Rightarrow 2^x\ge 1$

$\Rightarrow x\ge 0$

$letx<0$

$\Rightarrow {\log }_2({3.2}^{x-1}-1)<x$

$\Rightarrow {3.2}^{x-1}-1\le 2^x$

$\Rightarrow \frac{1}{2}{2}^x\le 1$

$\Rightarrow 2^x\le 1$

$\Rightarrow x\le 0$

$So,xϵR-\left\{0\right\}$

$and{3.2}^{x-1}-1>0$

$\Rightarrow 2^{x-1}>\frac{1}{3}$

$\Rightarrow 2^x>\frac{2}{3}$

$\Rightarrow x>lo{g}_2\left(\frac{2}{3}\right)$

$So,xϵ\left(lo{g}_2\left(\frac{2}{3}\right),\infty \right)$