Step 1: Understanding the Concept:
Logarithmic equations require a careful check of the domain. For \( \log_b(a) \), we must have \( a>0, b>0 \), and \( b \neq 1 \).
Step 2: Key Formula or Approach:
1. Determine the domain of \( \log_{x+1}(3 + 2x - x^2) \).
2. Simplify the LHS using the property \( b^{\log_b a} = a \).
3. Solve the resulting algebraic equation and check against the domain.
Step 3: Detailed Explanation:
Domain check:
1. Base: \( x + 1>0 \implies x>-1 \) and \( x + 1 \neq 1 \implies x \neq 0 \).
2. Argument: \( 3 + 2x - x^2>0 \implies x^2 - 2x - 3<0 \implies (x-3)(x+1)<0 \).
This gives \( x \in (-1, 3) \).
Combining these, the valid domain for the equation is \( x \in (-1, 3) \setminus \{0\} \).
LHS simplification:
In the interval \( x \in (-1, 3) \), \( x+1 \) is always positive, so \( |x+1| = x+1 \).
LHS \( = (x+1)^{\log_{x+1}(3 + 2x - x^2)} = 3 + 2x - x^2 \).
The equation becomes: \( 3 + 2x - x^2 = (x - 3)|x| \).
Case 1: \( x \in (0, 3) \). Then \( |x| = x \).
\( 3 + 2x - x^2 = x^2 - 3x \implies 2x^2 - 5x - 3 = 0 \).
Solving: \( (2x+1)(x-3) = 0 \).
Roots are \( x = -1/2 \) (not in this case's interval) and \( x = 3 \) (not in the domain as base must be strictly \(<3 \)).
Case 2: \( x \in (-1, 0) \). Then \( |x| = -x \).
\( 3 + 2x - x^2 = -x(x-3) = -x^2 + 3x \).
\( 3 + 2x = 3x \implies x = 3 \).
But this root \( x = 3 \) is not in the interval \( (-1, 0) \).
Step 4: Final Answer:
Since no solutions from the algebraic steps fall within the valid domain of the logarithmic function, the equation has no solution.