SAT 1000 problems - P9

If we perform a change-of-base operation on the latter logarithm function, we now obtain:

$f(x) = log_{2}(\sqrt{x}) \cdot \frac{log_{2}(2x)}{log_{2}(\sqrt{2})} = \frac{1}{2}log_{2}(x) \cdot \frac{log_{2}(2x)}{\frac{1}{2}} = log_{2}(x)[log_{2}(2) + log_{2}(x)] = [log_{2}(x)]^{2} + log_{2}(x).$

Taking the first derivative of $f(x)$ next produces:

$f'(x) = 0 \Rightarrow 2log_{2}(x) \cdot \frac{1}{x} + \frac{1}{x} = 0$ ;

or $log_{2}(x) = -\frac{1}{2} \Rightarrow x = \frac{1}{\sqrt{2}}$ .

A second-derivative check at $x = \frac{1}{\sqrt{2}}$ verifies that:

$f''(x) = \frac{1 - 2log_{2}(x)}{x^2} \Rightarrow f''(\frac{1}{\sqrt{2}}) = \frac{1 - 2log_{2}(\frac{1}{\sqrt{2}}) }{(\frac{1}{\sqrt{2}})^2} = \frac{1 - 2(-1/2)}{1/2} = 4 > 0$

hence, $f(x)$ attains its minimum value at $x = \frac{1}{\sqrt{2}}$ , or $f(\frac{1}{\sqrt{2}})$ = $\boxed{-\frac{1}{4}}.$