Question

If the roots of the quadratic equation $x^2-4x-{\log }_{3}a=0$ are real, then the least possible value of $a$ is

• A. $81$
• B. $\frac{1}{81}$
• C. $\frac{1}{64}$
• D. $9$

Answer 1

The correct option is B $\frac{1}{81}$

$x^2-4x-{\log }_{3}a=0$
Roots are real, so
$\Rightarrow D\ge 0$
$\Rightarrow 16+4{\log }_{3}a\ge 0$
$\Rightarrow {\log }_{3}a\ge -4$
As the base of the $\log$ is greater than $1$ the inequality will remain same, so
$\Rightarrow a\ge 3^{-4}$
$\Rightarrow a\ge \frac{1}{81}$

Hence, the least possible value of $a$ is $\frac{1}{81}$.