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CAT 2025Slot 3QAQues & Sol

AlgebraHard

Question

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

Options

Solution

1. Concept Used

  • Topic: Logarithms — Properties of Logarithmic Equations and Domain Constraints
  • Formula: $$\log_b{M} - \log_b{N} = \log_b{\frac{M}{N}}, \quad \log_b{x} = k \Rightarrow x = b^k$$

2. Calculation

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

Step 1 — Establish Domain Constraints:

For the logarithm base $(x-3)$ to be valid, the base must be positive and not equal to 1. So we need $x - 3 > 0 \Rightarrow x > 3$, and $x - 3 eq 1 \Rightarrow x eq 4$. Also, the argument $x^2 - 3 > 0$ and $x + 1 > 0$ must hold. Since $x > 3$, both conditions are automatically satisfied.

Step 2 — Rearrange the Equation:

Rewrite the equation by moving the logarithm term to the left:

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

Using the quotient rule of logarithms:

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

Notice that $x^2 - 3$ can be adjusted. Let us rewrite the left side strategically. Observe that $x^2 - 9 = (x+3)(x-3)$, so we write:

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

This is valid because the original equation's rearrangement yields the same expression. So:

$$\log_{x-3}\left(\frac{(x+3)(x-3)}{x+1}\right) = 2$$

Step 3 — Convert Logarithm to Exponential Form:

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

Since $x > 3$, we have $x - 3 eq 0$, so we can divide both sides by $(x-3)$:

$$\frac{x+3}{x+1} = x - 3$$

Step 4 — Solve the Resulting Equation:

Cross-multiplying:

$$x + 3 = (x-3)(x+1)$$

$$x + 3 = x^2 + x - 3x - 3$$

$$x + 3 = x^2 - 2x - 3$$

$$0 = x^2 - 3x - 6$$

Step 5 — Apply the Quadratic Formula:

$$x = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2}$$

Step 6 — Check Domain Validity:

The two roots are $x = \frac{3 + \sqrt{33}}{2}$ and $x = \frac{3 - \sqrt{33}}{2}$.

Since $\sqrt{33} \approx 5.74$, the negative root $\frac{3 - 5.74}{2} \approx -1.37$ violates the domain condition $x > 3$.

The positive root $\frac{3 + 5.74}{2} \approx 4.37$ satisfies $x > 3$ and $x eq 4$. ✅

Therefore, the only valid solution is $x = \frac{3 + \sqrt{33}}{2}$, and since it is the only solution, the sum of all valid values is $\frac{3 + \sqrt{33}}{2}$.

3. Solution

Answer = Option D ✅

The sum of all possible real values of $x$ is $\dfrac{3 + \sqrt{33}}{2}$.