Question

If the equation $x^2-4x+{\log }_{\frac{1}{2}}(a)=0$ does not have distinct real roots, then find the minimum value of $\frac{1}{a}$.

Answer 1

Answer: 16

$x^2-4x-{\log }_{2}a=0$
$D\le 0$
$16+4{\log }_{2}a\le 0\Rightarrow 4+{\log }_{2}a\le 0\Rightarrow {\log }_{2}a\le -4\Rightarrow a\le \frac{1}{16}$
${a_{\text{max. }}=\frac{1}{16}\Rightarrow \frac{1}{a}]}_{\min .}=16$