Question

The minimum value of x satisfying the inequality $4^{-x+0.5}-7\cdot 2^{-x}\le 4$ is

Answer 1

$\mathrm{Given}\mathrm{that}{4}^{-x+0.5}-7\cdot 2^{-x}\le 4....\left[1\right]\mathrm{Let}{2}^{-x}=t\mathrm{Now}\left[1\right]\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}2t^2-7t\le 4\Rightarrow 2t^2-7t-4\le 0\Rightarrow 2t^2-8t+t-4\le 0\Rightarrow 2t(t-4)+1(t-4)\le 0\Rightarrow \left(t-4\right)\left(2t+1\right)\le 0$


$\Rightarrow -\frac{1}{2}\le t\le 4\Rightarrow 0<t\le 4(Ast=2^{-x}and{2}^{-x}>0)\Rightarrow 0<2^{-x}\le 4\Rightarrow -2\le x<\infty \Rightarrow x\in [-2,\infty )$
The minimum value of the given expression is -2.