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

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Solution

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

$C a s e 1 : {log}_{10} x - 1 \geq 0 & e^{x} - 3 < 0$
$\Rightarrow {log}_{10} x \geq 1 & e^{x} < 3$
$\Rightarrow x \geq 10 & x < ln 3$
$∴ x \in ϕ$

$C a s e 2 : {log}_{10} x - 1 \leq 0 & e^{x} - 3 > 0$
$\Rightarrow {log}_{10} x \leq 1 & e^{x} > 3$
$\Rightarrow x \leq 10 & x > ln 3$
$∴ x \in (ln 3, 10]$
And $1 < l n 3 < 2 ∴ 2 \leq [x] \leq 10$
$[x]_{m i n} = 2 & [x]_{m a x} = 10 ∴ d i f f e r e n c e = 10 - 2 = 8.$

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