The domain of $$f(x) = \frac{\log_{(x+1)}(x-2)}{e^{2\log_e x^2 - (2x+3)}}$$, $$x \in R$$ is

To find the domain of the function, we need to determine the set of all real numbers $$x$$ for which both the numerator and the denominator are defined, where the denominator is not equal to zero.

$$f(x) = \frac{\log_{(x+1)}(x-2)}{e^{2\log_e x^2 - (2x+3)}}$$

Step 1: Analyze the Numerator

The numerator is a logarithmic function:

$$\log_{(x+1)}(x-2)$$

For a logarithm $$\log_b(a)$$ to be defined in real numbers, three conditions must be met:

The argument must be positive: $$x - 2 > 0 \implies x > 2$$

The base must be positive: $$x + 1 > 0 \implies x > -1$$

The base cannot equal 1: $$x + 1 \neq 1 \implies x \neq 0$$

Taking the intersection of these three conditions, the most restrictive constraint is:

$$x > 2$$

Any number greater than $$2$$automatically satisfies the conditions$$x > -1$$and$$x \neq 0$$.

Step 2: Analyze the Denominator

The denominator is an exponential function:

$$e^{2\log_e (x^2 - (2x+3))}$$

Logarithm in the exponent: The term $$\log_e (x^2 - (2x+3))$$ requires its argument to be strictly positive.

$$x^2 - (2x+3) > 0 \implies (x+1)(x-3) > 0 \implies x < -1$$ or $$x > 3$$

Non-zero denominator: The exponential function $$e^y$$ is always strictly greater than zero for any real number$$y$$. Therefore, the denominator will never equal zero.

Step 3: Combine the Conditions

To find the final domain, we take the intersection of the valid regions from Step 1, Step 2, and your added constraint:

Condition from the numerator: $$x > 2$$

Condition from the denominator: $$x < -1$$ or $$x > 3$$

The intersection of $$x > 2$$, $$x < -1$$ or $$x > 3$$ is:

$$x > 2 \text{ and } x \neq 3$$

Final Answer

The domain of the function is all real numbers strictly greater than $$2$$, excluding $$3$$. In interval notation, this is:

$$x \in (2, 3) \cup (3, \infty)$$