Question

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

• A. no solution
• B. two solutions
• C. unique solution
• D. infinite no. of solutions

Hint:

Always check the log domain constraints before doing any algebra! Here, notice how the argument constraint forces $x < 3$, while the right side contains a term $(x-3)$. This conflict at the boundary endpoint $x=3$ is a clear indicator that the roots are likely to be extraneous.

Answer 1

The Correct Option is A

Concept: Before attempting to solve a logarithmic equation algebraically, you must always establish its domain of existence constraints. For a log expression $\log_b(a)$ to be mathematically well-defined, the base and argument must satisfy: $$a > 0, \quad b > 0, \quad \text{and} \quad b \neq 1$$ Step 1: Establish the logarithmic domain constraints.
The given equation features the log term $\log_{x+1}(3+2x-x^2)$. Let let us write out the constraints for the base and argument:
• Base condition: $x + 1 > 0 \implies x > -1$
• Base exclusion condition: $x + 1 \neq 1 \implies x \neq 0$
• Argument condition: $3 + 2x - x^2 > 0 \implies x^2 - 2x - 3 < 0$ Factor the quadratic argument expression: $$(x - 3)(x + 1) < 0 \quad \Rightarrow \quad -1 < x < 3$$ Intersecting all three constraints gives the valid domain for the equation: $$\text{Domain: } x \in (-1, 3) \setminus \{0\} \quad \cdots (1)$$

Step 2: Simplify the logarithmic base equation.
Since $x > -1$, the term inside the absolute value brackets is strictly positive ($x+1 > 0$), meaning we can drop the bars: $|x+1| = x+1$. Now use the standard logarithmic identity $b^{\log_b(y)} = y$: $$(x+1)^{\log_{x+1}(3+2x-x^2)} = 3 + 2x - x^2$$ Substitute this simplified expression back into the main equation: $$3 + 2x - x^2 = (x - 3)|x| \quad \cdots (2)$$

Step 3: Solve the algebraic equation within our domain branches.
Let us analyze equation (2) across the two separate branches of our domain from equation (1):
• Branch 1: For $x \in (-1, 0)$: Here, $x$ is negative, so $|x| = -x$. Substitute this into equation (2): $$3 + 2x - x^2 = (x - 3)(-x) \quad \Rightarrow \quad 3 + 2x - x^2 = -x^2 + 3x$$ Cancel out $-x^2$ from both sides: $$3 + 2x = 3x \quad \Rightarrow \quad x = 3$$ However, the value $x = 3$ lies completely outside this branch interval $(-1, 0)$. Thus, it is an extraneous root.
• Branch 2: For $x \in (0, 3)$: Here, $x$ is positive, so $|x| = x$. Substitute this into equation (2): $$3 + 2x - x^2 = (x - 3)(x) \quad \Rightarrow \quad 3 + 2x - x^2 = x^2 - 3x$$ Bring all terms to the right side to form a quadratic equation: $$2x^2 - 5x - 3 = 0 \quad \Rightarrow \quad (2x + 1)(x - 3) = 0$$ This yields the roots $x = -\frac{1}{2}$ and $x = 3$. Let us verify if either root falls inside our current branch interval $(0, 3)$: - $x = -\frac{1}{2}$ is negative, so it is excluded from $(0, 3)$. - $x = 3$ is an open boundary endpoint, so it is also excluded from the open interval $(0, 3)$.

Step 4: Conclude the total number of solutions.
Since every algebraically derived root fails to satisfy the logarithmic domain constraints, the equation has no valid real solution, corresponding to choice (A).