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

Solution :

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

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

(the sign of the inequality is retained)

Now x + 2 + $$2\over x$$  $$\implies$$ $$x^2 + 2x + 2\over x$$ > 0

$$\implies$$ $$(x + 1)^2 + 1\over x$$ > 0  $$\implies$$  x $$\in$$ $$(0, \infty)$$

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