Question

Solve the following inequality:
$4^x-3⩽2^{x+1}$

Answer 1

${{(2}^x)}^2-3\le 2.2^x$

$let{2}^x=t(t>0)$

$t^2-3\le 2t$

$t^2-2t-3\le 0$

$t^2-3t+t-3\le 0$

$(t-3)(t+1)\le 0$

$t\epsilon (-1,3)$

$butt>0,t\epsilon (0,3)$

$2^x\epsilon (0,3)$

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