Question

If $\frac{({\log }_{10}x-1)}{(e^x-3)}\le 0$, then difference between greatest and smallest integral value satisfying the inequality is

Answer 1

$\because \frac{({\log }_{10}x-1)}{(e^x-3)}\le 0,$ which is possible in following two cases.

$Case1:{\log }_{10}x-1\ge 0\&e^x-3<0$
$\Rightarrow {\log }_{10}x\ge 1\&e^x<3$
$\Rightarrow x\ge 10\&x<\ln 3$
$\therefore x\in \varphi$

$Case2:{\log }_{10}x-1\le 0\&e^x-3>0$
$\Rightarrow {\log }_{10}x\le 1\&e^x>3$
$\Rightarrow x\le 10\&x>\ln 3$
$\therefore x\in (\ln 3,10]$
And $1<ln3<2\therefore 2\le \left[x\right]\le 10$
$[x{]}_{min}=2\&[x{]}_{max}=10\therefore difference=10-2=8.$