Let [x] denote the greatest integer less than or equal

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

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

Let $\left[e^{x}\right]=t$

$\Rightarrow \mathrm{t}^{2}+\mathrm{t}-2=0$

$\Rightarrow \mathrm{t}=-2,1$

$\left[e^{x}\right]=-2$ (Not possible)

or $\left[e^{x}\right]=1 \quad \therefore 1 \leq e^{x}<2$

$\Rightarrow \quad \ln (1) \leq x<\ln (2)$

$\Rightarrow \quad 0 \leq x<\ln (2)$

$\Rightarrow \quad x \in[0, \ln 2)$