Let [x] denote the greatest integer less than or equal to x . Then,...

Given equation is ${\left[{e}^{x}\right]}^{2} + \left[{e}^{x} + 1\right] - 3 = 0$

We know that \left[x + I\right] = \left[x\right] + I, I \in Z,$where \left[\cdot\right],$ represents the greatest integer function.

$⇒ {\left[{e}^{x}\right]}^{2} + \left[{e}^{x}\right] + 1 - 3 = 0$

Let, \left[{e}^{x}\right] = t$</p><p>$⇒ {t}^{2} + t - 2 = 0$</p><p>$⇒ \left(t + 2\right)\left(t - 1\right) = 0$</p><p>$⇒ t = - 2, 1$</p><p>But \left[{e}^{x}\right] = - 2, ($Not possible because we know that ${e}^{x} > 0 )$

Hence, \left[{e}^{x}\right] = 1 and also, we know that if $\left[x\right] = 1, ⇒ 1 \le x < 2$

$∴ 1 \le {e}^{x} < 2$

And, now by using the definition of logarithm, we have, if ${e}^{x} = a,$ then $\log_{e}a = x$

$⇒ \log_{e}1 \le x < \log_{e}2$

$⇒ 0 \le x < \log_{e}2$

$⇒ x \in \left[0, \log_{e}2\right) .</p>$