CAT 2025 Slot 3 · Quant · Question 7

A logarithm needs a positive base that is not 1, so x-3>0 and x-3\ne 1, giving x>3 and x\ne 4.

Bringing the two logarithms together, {\log}_{x-3}\left( \frac{{x}^{2}-9}{x+1}\right)=2.

Removing the logarithm, \frac{{x}^{2}-9}{x+1}={\left( x-3 \right)}^{2}.

Factorising the numerator, \frac{\left( x-3 \right)\left( x+3 \right)}{x+1}={\left( x-3 \right)}^{2}.

Since x>3, the factor x-3 is non-zero and can be cancelled: \frac{x+3}{x+1}=x-3.

Cross-multiplying, x+3=\left( x-3 \right)\left( x+1 \right)={x}^{2}-2x-3, so {x}^{2}-3x-6=0.

The roots are x=\frac{3\pm \sqrt{9+24}}{2}=\frac{3\pm \sqrt{33}}{2}.

Since \sqrt{33}\approx 5.74, the negative root is about -1.37, which fails x>3; the other root is about 4.37, which satisfies x>3 and x\ne 4.

Only one value survives, so the sum of all possible values is \frac{3+\sqrt{33}}{2}.

Shortcut. Check the domain before solving. With x>3 forced, at most one root of a quadratic with a negative product of roots can survive — so the sum of valid values is just that root, and the only option exceeding 3 is the answer.