1. Key Concepts and Formulas

• Greatest Integer Function: $[x]$ denotes the greatest integer less than or equal to $x$. For any integer $n$, $[x+n] = [x] + n$.
• Exponential and Logarithmic Functions: $e^x$ is always positive for real $x$. $\ln(x)$ is the natural logarithm (base $e$). $\ln(e^x) = x$ and $e^{\ln(x)} = x$. $\ln(1) = 0$.
• Definition of Greatest Integer Function: $[x] = n$ if and only if $n \le x < n+1$, where $n$ is an integer.

2. Step-by-Step Solution

Step 1: Simplify the equation using the greatest integer function property.

We are given the equation ${[{e^x}]^2} + [{e^x} + 1] - 3 = 0$. We use the property $[x+n] = [x] + n$ to simplify the second term. Since $[{e^x} + 1] = [{e^x}] + 1$, the equation becomes: ${[{e^x}]^2} + [{e^x}] + 1 - 3 = 0$ ${[{e^x}]^2} + [{e^x}] - 2 = 0$ Why: This simplifies the equation and makes it easier to manipulate into a quadratic form.

Step 2: Substitute to create a quadratic equation.

Let $t = [{e^x}]$. The equation then becomes: $t^2 + t - 2 = 0$ Why: This substitution transforms the original equation into a standard quadratic equation, which is easier to solve.

Step 3: Solve the quadratic equation for t.

The quadratic equation $t^2 + t - 2 = 0$ can be factored as: $(t+2)(t-1) = 0$ Thus, the solutions are $t = -2$ or $t = 1$. Why: Solving for $t$ gives us the possible values of $[{e^x}]$.

Step 4: Validate the solutions for t.

Since $e^x > 0$ for all real $x$, it follows that $[{e^x}]$ must be a non-negative integer. Therefore, $t$ must be a non-negative integer.

• If $t = -2$, then $[{e^x}] = -2$. This is impossible since $e^x > 0$ and thus $[e^x]$ must be non-negative.
• If $t = 1$, then $[{e^x}] = 1$. This is a valid solution.

Why: We must check if the solutions are valid within the context of the original problem. Since $e^x$ is always positive, its greatest integer cannot be negative.

Step 5: Determine the range of $e^x$.

Since $[{e^x}] = 1$, we know that $1 \le e^x < 2$ Why: This inequality represents all possible values of $e^x$ such that its greatest integer is 1.

Step 6: Solve for x using logarithms.

To find the range of $x$, we take the natural logarithm of all parts of the inequality: $\ln(1) \le \ln(e^x) < \ln(2)$ Since $\ln(1) = 0$ and $\ln(e^x) = x$, we have: $0 \le x < \ln(2)$ Why: Applying the natural logarithm allows us to isolate $x$ and determine its range.

Step 7: Match the solution with the given options.

The solution is $0 \le x < \ln(2)$, which means $x \in [0, \ln(2))$. Since $\ln(2) = \log_e 2$, this interval is $[0, \log_e 2)$. This matches option (D).

Why: The final step is to verify that the derived solution matches one of the given options.

3. Common Mistakes & Tips

• Forgetting to validate the solutions for t: Always check if the solutions obtained after solving the quadratic equation are valid in the context of the original problem. In this case, since $e^x > 0$, then $[e^x]$ must be non-negative.
• Incorrectly applying the greatest integer function definition: Remember that if $[x] = n$, then $n \le x < n+1$.
• Logarithm Errors: Be careful when applying logarithm properties, especially when dealing with inequalities.

4. Summary

The problem is solved by using the properties of the greatest integer function to simplify the given equation. Then, we make a substitution to transform the equation into a quadratic equation. After solving the quadratic equation, we check the validity of the solutions and use the definition of the greatest integer function and logarithms to determine the range of $x$. The values of $x$ satisfying the equation lie in the interval $[0, \ln(2))$.

5. Final Answer

The final answer is \boxed{[0, \log_e 2)}, which corresponds to option (D).