Solution of 2^x+2^ |x| ≥ 2 2 is

2^x+2^{|x|} \geq 2 \sqrt{2}$...(i)<br />Case I:$x \geq 0$, then Eq. (i) becomes<br />$2^x+2^x \geq 2 \sqrt{2}$<br />$\Rightarrow 2^x \geq \sqrt{2} \Rightarrow x \geq \frac{1}{2}$<br />Case II:$x < 0$, then Eq. (i) becomes<br />$2^x+2^{-x} \geq 2 \sqrt{2}$<br />$\Rightarrow t+\frac{1}{t} \geq 2 \sqrt{2}$, where$2^x=t$<br />$\Rightarrow t^2-2 \sqrt{2} t+1 \geq 0$<br />$\Rightarrow x \leq \log _2(\sqrt{2}-1)$<br />Also,$0 < \sqrt{2}-1 < 1, \log _2(\sqrt{2}-1) < 0$.<br />$\thereforeThe solution is<br />\left.\left(-\infty, \log _ _2(\sqrt{2}- 1)\right)\right] \cup\left[\frac{1}{2}, \infty\right)