The equation $|x + 1|^{\log_{(x+1)}(3 + 2x - x^2)} = (x - 3)|x|$ has

For the given equation to be defined, the base and the argument of the logarithm must satisfy specific conditions.

Base of the logarithm: $x + 1 > 0$ and $x + 1 \neq 1$
$\Rightarrow x > -1$ and $x \neq 0$

Argument of the logarithm: $3 + 2x - x^2 > 0$
$\Rightarrow x^2 - 2x - 3 < 0$
$\Rightarrow (x - 3)(x + 1) < 0$

Since $x > -1$, we know $x + 1 > 0$. Therefore, we must have $x - 3 < 0 \Rightarrow x < 3$.
Thus, the domain of the equation is $x \in (-1, 0) \cup (0, 3)$.

In this domain, $x + 1 > 0$, so $|x + 1| = x + 1$. The equation simplifies to:
$(x + 1)^{\log_{(x+1)}(3 + 2x - x^2)} = (x - 3)|x|$

Using the logarithmic identity $a^{\log_a b} = b$, the left-hand side becomes $3 + 2x - x^2$.
$\Rightarrow 3 + 2x - x^2 = (x - 3)|x|$
$\Rightarrow -(x^2 - 2x - 3) = (x - 3)|x|$
$\Rightarrow -(x - 3)(x + 1) = (x - 3)|x|$

Since $x < 3$, we know $x - 3 \neq 0$. Dividing both sides by $(x - 3)$, we get:
$-(x + 1) = |x|$

Now, we analyze this in the two intervals of our domain:

Case 1: $x \in (0, 3)$
Here, $|x| = x$. The equation becomes:
$-(x + 1) = x$
$\Rightarrow -x - 1 = x$
$\Rightarrow 2x = -1 \Rightarrow x = -\dfrac{1}{2}$
This value is not in the interval $(0, 3)$, so it is rejected.

Case 2: $x \in (-1, 0)$
Here, $|x| = -x$. The equation becomes:
$-(x + 1) = -x$
$\Rightarrow -x - 1 = -x$
$\Rightarrow -1 = 0$
This is a contradiction, yielding no solution.

Since both cases yield no valid values for $x$, the equation has no solution.

Answer: no solution