Let [x] denote the greatest integer less than | vedclass

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(C) Given the equation: $[e^{x}]^{2} + [e^{x} + 1] - 3 = 0$. Using the property $[x + n] = [x] + n$ for any integer $n$,we have $[e^{x} + 1] = [e^{x}]

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$[0, \log_{e} 2)$

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(C) Given the equation: $[e^{x}]^{2} + [e^{x} + 1] - 3 = 0$. Using the property $[x + n] = [x] + n$ for any integer $n$,we have $[e^{x} + 1] = [e^{x}] + 1$. Substituting this into the equation: $[e^{x}]^{2} + [e^{x}] + 1 - 3 = 0$. Let $t = [e^{x}]$. Then the equation becomes $t^{2} + t - 2 = 0$. Factoring the quadratic: $(t + 2)(t - 1) = 0$,which gives $t = -2$ or $t = 1$. Since $e^{x} > 0$,$[e^{x}]$ cannot be $-2$. Thus,$[e^{x}] = 1$. By the definition of the greatest integer function,$1 \leq e^{x} Taking the natural logarithm on all sides: $\ln(1) \leq x Since $\ln(1) = 0$,we get $0 \leq x Therefore,$x \in [0, \ln 2)$.

Let $[x]$ denote the greatest integer less than or equal to $x$. Then,the values of $x \in \mathbb{R}$ satisfying the equation $[e^{x}]^{2} + [e^{x} + 1] - 3 = 0$ lie in the interval:

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