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,\(\infty\)).