Question

If $\frac{{\left({\log }_{e}x\right)}^2-3{\log }_{e}x+3}{{\log }_{e}x-1}<1$, then

• A. $x\in (0,e)$
• B. $x\in (1,e)$
• C. $x\in (e,1)$
• D. none of these

Answer 1

The correct option is A $x\in (0,e)$

$\frac{{\left({\log }_{e}x\right)}^2-3{\log }_{e}x+3}{{\log }_{e}x-1}$ is valid, when $x>0$
$\frac{{\left({\log }_{e}x\right)}^2-3{\log }_{e}x+3}{{\log }_{e}x-1}<1$
$\Rightarrow \frac{{\left({\log }_{e}x\right)}^2-3{\log }_{e}x+3}{{\log }_{e}x-1}-1<0$
$\Rightarrow \frac{{\left({\log }_{e}x\right)}^2-4{\log }_{e}x+4}{{\log }_{e}x-1}<0$
$\Rightarrow \frac{{({\log }_{e}x-2)}^2}{{\log }_{e}x-1}<0$
$\Rightarrow {\log }_{e}x<1$
Since, $x>0$
Therefore, $x\in (0,e)$
Ans: A