1. Key Concepts and Formulas

• Logarithmic Function Domain Restrictions: For $\log_b a$ to be defined, we require $a > 0$, $b > 0$, and $b \neq 1$.
• Rational Function Domain Restrictions: For a fraction $\frac{P(x)}{Q(x)}$ to be defined, the denominator $Q(x)$ must not be equal to zero, i.e., $Q(x) \neq 0$.
• Exponential Function Simplification: $e^{k \log_e x} = e^{\log_e x^k} = x^k$.

2. Step-by-Step Solution

The function is given by $f(x) = {{{{\log }_{(x + 1)}}(x - 2)} \over {{e^{2{{\log }_e}x}} - (2x + 3)}}$. To find the domain, we need to ensure that all parts of the function are well-defined.

Step 1: Analyze the numerator $\log_{(x+1)}(x-2)$ For the logarithmic term $\log_{(x+1)}(x-2)$ to be defined, we must satisfy three conditions:

• Argument is positive: $x - 2 > 0 \implies x > 2$.
• Base is positive: $x + 1 > 0 \implies x > -1$.
• Base is not equal to one: $x + 1 \neq 1 \implies x \neq 0$.

Combining these conditions, we need $x > 2$.

Step 2: Analyze the denominator $e^{2{{\log }_e}x} - (2x + 3)$ First, let's simplify the term $e^{2{{\log }_e}x}$. Using the logarithm property $k \log_e x = \log_e x^k$, we get $2 \log_e x = \log_e x^2$. Then, using the property $e^{\log_e y} = y$, we have $e^{2{{\log }_e}x} = e^{\log_e x^2} = x^2$.

For the term $e^{2{{\log }_e}x}$ to be defined, the argument of the logarithm, $x$, must be positive. So, $x > 0$.

Now, the denominator becomes $x^2 - (2x + 3)$. For the entire function to be defined, the denominator cannot be zero. So, we must have $x^2 - (2x + 3) \neq 0$. $x^2 - 2x - 3 \neq 0$. Factoring the quadratic expression: $(x - 3)(x + 1) \neq 0$. This implies $x \neq 3$ and $x \neq -1$.

Step 3: Combine all domain restrictions We have the following restrictions from the numerator and the denominator:

• From the numerator's logarithm: $x > 2$.
• From the denominator's exponential term: $x > 0$.
• From the denominator being non-zero: $x \neq 3$ and $x \neq -1$.

To satisfy all these conditions simultaneously, we need to find the intersection of the intervals and excluded points. The most restrictive condition from the numerator is $x > 2$. The condition $x > 0$ from the exponential term is already covered by $x > 2$. The excluded points are $x \neq 3$ and $x \neq -1$. The condition $x > 2$ already excludes $x = -1$. Therefore, we need $x > 2$ and $x \neq 3$.

This can be written in interval notation as $(2, \infty) - \{3\}$.

3. Common Mistakes & Tips

• Forgetting the base conditions of logarithms: Always remember that the base of a logarithm must be positive and not equal to one.
• Confusing argument and base: Ensure you correctly apply the conditions to the argument and the base of the logarithm.
• Simplifying too early without considering domain: While simplifying $e^{2 \log_e x}$ to $x^2$ is correct, remember that the original expression $2 \log_e x$ requires $x > 0$.

4. Summary

To determine the domain of the given function, we analyzed the constraints imposed by the logarithmic terms and the rational expression. The numerator's logarithmic term $\log_{(x+1)}(x-2)$ requires the argument $x-2 > 0$, the base $x+1 > 0$, and the base $x+1 \neq 1$. These conditions lead to $x > 2$. The denominator's term $e^{2 \log_e x}$ requires $x > 0$. Furthermore, the denominator itself cannot be zero, so $x^2 - (2x+3) \neq 0$, which means $x \neq 3$ and $x \neq -1$. Combining all these conditions, the most stringent requirement is $x > 2$, along with the exclusion of $x=3$. Thus, the domain is $(2, \infty) - \{3\}$.

5. Final Answer

The final answer is $\boxed{(2,\infty ) - \{ 3\} }$, which corresponds to option (C).