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 restrictions imposed by both the numerator and the denominator.
The numerator is \(\log_{x+1}(x-2)\). For a logarithm \(\log_b(a)\) to be defined:
\(b = x + 1 > 0 \Rightarrow x > -1\)
\(b \neq 1 \Rightarrow x + 1 \neq 1 \Rightarrow x \neq 0\)
\(a = x - 2 > 0 \Rightarrow x > 2\)
Next, consider the denominator \(e^{2\log x}-(2x+3)\). This expression must not be zero:
\(e^{2\log x} = (e^{\log x})^2 = x^2\). So, we have \(x^2 \neq 2x + 3\).
This can be rewritten as \(x^2 - 2x - 3 \neq 0\).
Solving \(x^2 - 2x - 3 = 0\), we factor it as \((x-3)(x+1)=0\).
Hence, the solutions are \(x = 3\) and \(x = -1\).
We exclude these values, taking into account \(x \neq 3\) since \(x > 2\).
Taking all these restrictions into account, we combine the conditions to find the domain:
\(x > 2\) from the conditions on the numerator.
\(x \neq 3\) from the conditions on the denominator.
Therefore, the domain of the function is \(x \in (2, \infty) - \{3\}\).
