Question

The solution set of ${\log }_{1/2}(2^{x+2}-4^x)\ge -2$ is

Answer 1

${\log }_{y_2}(2^{x+2}-4^x)\ge -2$

$\therefore -{\log }_2(2^{x+2}-4^x)\ge -2$

$\therefore -{\log }_2(2^{x+2}-4^x)\le -2$

$\therefore 2^{x+2}-4^x\le 4$

$\therefore 4.x{2}^2-2^{2x}\le 4$

Let $2^x=t$

$\therefore 4t-t^2$

$\therefore 0\le t^2-4t+4$

$\therefore 0\le (t-2{)}^2$

$H$ values