Question

Solve the following inequality:
$2^{2x+1}-21\cdot {\left(\frac{1}{2}\right)}^{2x+3}+2⩾0$

Answer 1

$2^{2x+1}-21.{\left(\frac{1}{2}\right)}^{2x+3}+2\ge 0$

$2.{{(2}^x)}^2-21.{\frac{2}{8}}^{-2x}+2\ge 0$

$2.2^x-\frac{21}{8}{2}^{-2x}+2\ge 0$

$let{2}^{2x}=t(t>0)2t-\frac{21}{8t}+2\ge 0$

$16t^2-21+16t\ge 0$

$16t^2+16t-21\ge 0$

$t^2+t-\frac{21}{6}\ge 0$

$(t-\frac{3}{4})(t+\frac{7}{4})\ge 0$

$t\epsilon (-\infty ,\frac{-7}{4})\bigcup (\frac{3}{4},\infty )$

$butt>0$

$t\epsilon (\frac{3}{4},\infty )$

${{(2}^x)}^2\epsilon (\frac{3}{4},\infty )$

$2^x\epsilon (\frac{\sqrt{3}}{2},\infty )$

$x\epsilon (\frac{\log \frac{\sqrt{3}}{2}}{\log 2},\infty )$