Question

If $\mid {\log }_{3}x\mid -{\log }_{3}x-3<0$ and the value of $x$ lies in the integral $(\frac{1}{\sqrt{k}},\infty )$, then the value of $(k{)}^{\frac{1}{3}}$ is

Answer 1

$\mid {\log }_{3}x\mid -{\log }_{3}x-3<0$
When $x\ge 1$,
${\log }_{3}x-{\log }_{3}x-3<0$
$\Rightarrow -3<0$
$x\in [1,\infty ).....\left(1\right)$
When $x<1$,
$-{\log }_{3}x-{\log }_{3}x-3<0$
$\Rightarrow -2{\log }_{3}x-3<0$
$\Rightarrow {\log }_3{x}^{-2}-3<0,[\because a\log b=\log b^a]$
$\Rightarrow {\log }_3\frac{1}{x^2}<3$
$\Rightarrow \frac{1}{x^2}<3^3=27$
$\Rightarrow x\in (\frac{1}{\sqrt{27}},1).....\left(2\right)$
So, from $\left(1\right)$ and $\left(2\right)$,
$x\in (\frac{1}{\sqrt{27}},\infty )$
$\Rightarrow k=27$
$\Rightarrow k^{\frac{1}{3}}=3$.