Question

Solve the following inequality:
$4^{-x+0.5}-7\cdot 2^{-x}-4<0$

Answer 1

$4^{-x+0.5}-7(2{)}^{-x}-4<0$

$\frac{\sqrt{4}}{4^x}-\frac{7}{2^x}-4<0$

$\frac{2}{{(2}^x{)}^2}-\frac{7}{2^x}-4<0$

$let{2}^x=t$

$\frac{2}{t^2}-\frac{7}{t}-4<0$

$2-7t-4t^2<0$

$4t^2+7t-2>0$

$4t^2+8t-t-2>0$

$(4t-1)(t+2)>0$

$(t-\frac{1}{4})(t+2)>0$

$t\epsilon (-\infty ,-2)\bigcup (\frac{1}{4},\infty )$

$butt>0hence$

$t\epsilon (\frac{1}{4},\infty )$

$2^x\epsilon (\frac{1}{4},\infty )$

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