# Solve for x : $$2^{x+2}$$ > $$({1\over 4})^{1/x}$$

## Solution :

We have $$2^{x+2}$$ > $$2^{-2/x}$$

Since the base 2 > 1, we have x + 2 > $$-2\over x$$

(the sign of inequality is retained)

Now, x + 2 + $$-2\over x$$ > 0 $$\implies$$ $${x^2 + 2x + 2}\over x$$ > 0

$$\implies$$  $$({x+1})^2 + 1\over x$$ > 0

$$\implies$$  x $$\in$$ (0,$$\infty$$).