The domain of `f(x)=(log_(x+1)(x-2))/(e^(2log x)-(2x+3)),x in R` is

Options

A. `(2,oo)-{3}`
B. `R-{3}`
C. `R-{-1,3}`
D. `(-1,oo)-{3}`

Answer

To find the domain of the function \( f(x) = \frac{\log_{(x+1)}(x-2)}{e^{2\log x} - (2x + 3)} \), we need to consider the conditions under which both the numerator and denominator are defined and valid.

### Step 1: Analyze the numerator

The numerator is \( \log_{(x+1)}(x-2) \).

1. **Base Condition**: The base \( x + 1 \) must be greater than 0 and not equal to 1:
- \( x + 1 > 0 \) implies \( x > -1 \)
- \( x + 1 \neq 1 \) implies \( x \neq 0 \)

2. **Argument Condition**: The argument \( x - 2 \) must be greater than 0:
- \( x - 2 > 0 \) implies \( x > 2 \)

### Step 2: Combine conditions from the numerator

From the numerator, we have:
- \( x > 2 \)
- \( x > -1 \) (this is automatically satisfied if \( x > 2 \))
- \( x \neq 0 \) (this is also automatically satisfied if \( x > 2 \))

Thus, the conditions from the numerator simplify to:
- \( x > 2 \)

### Step 3: Analyze the denominator

The denominator is \( e^{2\log x} - (2x + 3) \).

1. **Logarithm Condition**: The logarithm \( \log x \) must be defined:
- \( x > 0 \)

2. **Exponential Simplification**: We can simplify \( e^{2\log x} \):
- \( e^{2\log x} = (e^{\log x})^2 = x^2 \)

Thus, the denominator becomes:
- \( x^2 - (2x + 3) \neq 0 \)
- Simplifying this gives \( x^2 - 2x - 3 \neq 0 \)

3. **Factoring the quadratic**:
- \( x^2 - 2x - 3 = (x - 3)(x + 1) \)
- Therefore, \( (x - 3)(x + 1) \neq 0 \) implies \( x \neq 3 \) and \( x \neq -1 \)

### Step 4: Combine conditions from the denominator

From the denominator, we have:
- \( x > 0 \)
- \( x \neq 3 \)

### Step 5: Final domain determination

Now we combine all the conditions obtained from both the numerator and denominator:
- From the numerator: \( x > 2 \)
- From the denominator: \( x > 0 \) and \( x \neq 3 \)

Since \( x > 2 \) already satisfies \( x > 0 \), we only need to exclude \( x = 3 \).

Thus, the domain of \( f(x) \) is:
\[ \text{Domain: } (2, 3) \cup (3, \infty) \]

### Final Answer:
The domain of \( f(x) \) is \( (2, 3) \cup (3, \infty) \).