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CAT 2025Slot 3QAQues & Sol

AlgebraMedium

Question

The sum of all possible real values of $x$ satisfying $\log_{x-3}(x^2-9) = \log_{x-3}(x+1) + 2$ is

Options

Solution

1. Concept Used

  • Topic: Logarithms — Properties of Logarithmic Equations and Domain Constraints
  • Key idea: Express the constant 2 as $\log_{x-3}(x-3)^2$, apply the quotient rule, then equate arguments.

2. Calculation

Step 1 — Domain Constraints:

Base $(x-3)$ requires $x-3>0 \Rightarrow x>3$ and $x-3 eq 1 \Rightarrow x eq 4$. For $x>3$, both arguments $x^2-9>0$ and $x+1>0$ are automatically satisfied.

Step 2 — Rearrange:

Rewrite $2 = \log_{x-3}(x-3)^2$ and move the log terms together:

$$\log_{x-3}\frac{x^2-9}{x+1} = \log_{x-3}(x-3)^2$$

Equating arguments:

$$\frac{(x+3)(x-3)}{x+1} = (x-3)^2$$

Step 3 — Simplify and Solve:

Divide both sides by $(x-3)$, valid since $x>3$:

$$\frac{x+3}{x+1} = x-3$$ $$x+3 = (x-3)(x+1) = x^2-2x-3$$ $$x^2-3x-6=0$$ $$x = \frac{3\pm\sqrt{33}}{2}$$

Step 4 — Check Domain:

  • $x = \frac{3+\sqrt{33}}{2}\approx 4.37$: satisfies $x>3$ and $x eq 4$ ✓
  • $x = \frac{3-\sqrt{33}}{2}\approx -1.37$: violates $x>3$ ✗

The only valid solution is $\dfrac{3+\sqrt{33}}{2}$, so the sum of all valid values is $\dfrac{3+\sqrt{33}}{2}$.

3. Answer

Answer = Option D ✅