Question

Solve the following inequality:
${\log }_{1/3}(2^{x+2}-4^x)⩾-2$

Answer 1

$let{2}^x=z$

$-{\log }_3(4z-z^2)\ge -2$

$4z-z^2\le 9$

$z^2-4z+9\ge 0$

$thereforez\epsilon R$

$now,2^{x+2}-4^x>0$

$as{2}^x=z$

$4z-z^2>0$

$z^2-4z<0$

$z(z-4)<0$

$z\epsilon (0,4)$

$2^x\epsilon (0,4)$

$x\epsilon (-\infty ,2)$