Question

Solve the following inequality:
${\left(\frac{1}{3}\right)}^{\sqrt{x+2}}<3^{-x}$

Answer 1

${\left(\frac{1}{3}\right)}^{\sqrt{x+2}}<3^{-x}\Rightarrow 3^{-\sqrt{x+2}}<3^{-x}\Rightarrow -\sqrt{x+2}<-x\Rightarrow \sqrt{x+2}>x$

For $\sqrt{x+2}$ to be valid, $x+2\ge 0\Rightarrow x\ge -2$

Therefore, $\sqrt{x+2}$ will be equal to $x$ at $x=2$

Therefore, $\sqrt{x+2}=x$ at $x<2$

Therefore, $x\in (-2,2)$