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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