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

The function is:

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

We need to evaluate the constraints for both the numerator and the denominator independently, and then find their intersection.

1. Constraints for the Numerator

For the logarithmic function $$\log_{(x+1)}(x-2)$$ to be defined, the following conditions must be met:

• The argument must be strictly positive:

  $$x - 2 > 0 \implies x > 2$$

• The base must be strictly positive:

  $$x + 1 > 0 \implies x > -1$$

• The base cannot equal 1:

  $$x + 1 \neq 1 \implies x \neq 0$$

Intersecting these three conditions for the numerator, we get $$x > 2$$.

2. Constraints for the Denominator

The denominator must not equal zero, and any functions within it must also be defined.

• The argument of the logarithm in the exponent must be strictly positive:

  $$x > 0$$

• The denominator cannot be zero:

  $$e^{2\log_e x} - (2x+3) \neq 0$$

Using the logarithm power rule, $$2\log_e x = \log_e(x^2)$$. For $$x > 0$$, the expression simplifies using the property $$e^{\log_e(a)} = a$$:

$$e^{\log_e(x^2)} = x^2$$

Substitute this back into the denominator inequality:

$$x^2 - (2x + 3) \neq 0$$
$$x^2 - 2x - 3 \neq 0$$

Factoring the quadratic equation gives:

$$(x - 3)(x + 1) \neq 0$$

This means $$x \neq 3$$ and $$x \neq -1$$.

3. Final Domain

To find the domain of $$f(x)$$, we take the intersection of the valid sets for both the numerator and the denominator:

• From the numerator: $$x > 2$$
• From the denominator: $$x > 0$$, $$x \neq 3$$, and $$x \neq -1$$

Combining these constraints, $$x$$ must be strictly greater than $$2$$, but it cannot equal $$3$$.

Therefore, the domain of $$f(x)$$ is:

$$x \in (2, 3) \cup (3, \infty)$$ or $$(2, \infty) \setminus \{3\}$$.