Find the range \(log_4(x + \frac 1x)\) .

1.	Find the range \(log_4(x + \frac 1x)\) .
Answer» \(f(x) = log_4(x + \frac 1x)\) Domain of f(x), \(x + \frac 1x > 0\) ⇒ \(\frac{x^2 + 1}x > 0\) ⇒ \(x \in (0, \infty)\) \((\because x^2 + 1 > 0)\) \(\because x + \frac 1x = (\sqrt x)^2 + (\frac 1{\sqrt x})^2 - 2\sqrt x \times \frac 1{\sqrt x} + 2\) \(= (\sqrt x - \frac1{\sqrt x})^2 + 2\) \(\ge 2\) \(\therefore log_4 (x + \frac 1x) \ge log_42\) (\(\because\) log is an increasing function) ⇒ \(f(x) \ge \frac 12 \) \(\left(\because log_42 = \frac{log\,2}{log\,4} = \frac{log\,2}{2log\,2} = \frac12\right)\) ⇒ \(f(x) \in [0.5, \infty)\).

Answer»

\(f(x) = log_4(x + \frac 1x)\)

Domain of f(x), \(x + \frac 1x > 0\)

⇒ \(\frac{x^2 + 1}x > 0\)

⇒ \(x \in (0, \infty)\) \((\because x^2 + 1 > 0)\)

\(\because x + \frac 1x = (\sqrt x)^2 + (\frac 1{\sqrt x})^2 - 2\sqrt x \times \frac 1{\sqrt x} + 2\)

\(= (\sqrt x - \frac1{\sqrt x})^2 + 2\)

\(\ge 2\)

\(\therefore log_4 (x + \frac 1x) \ge log_42\) (\(\because\) log is an increasing function)

⇒ \(f(x) \ge \frac 12 \) \(\left(\because log_42 = \frac{log\,2}{log\,4} = \frac{log\,2}{2log\,2} = \frac12\right)\)

⇒ \(f(x) \in [0.5, \infty)\).

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