CAT 2025 Slot 2 QA Q5 Logarithms & Exponents

Question 5 of the Quantitative Ability (QA) section, CAT 2025 Slot 2, with CAT's official answer and a worked solution.

If \( 9 ^ { x ^ { 2 } + 2 x - 3 } - 4 \left( 3 ^ { x ^ { 2 } + 2 x - 2 } \right) + 27 = 0 \), then the product of all possible values of \( x \) is

A
\( 5 \)
B
\( 20 \)
C
\( 30 \)
D
\( 15 \)

Answer: (B) 20 Logarithms & Exponents MCQ

Solution

Let \(t = 3^{x^2+2x-3}\). Then \(9^{x^2+2x-3} = t^2\) and \(3^{x^2+2x-2} = 3t\).
Equation: \(t^2 - 12t + 27 = 0 \Rightarrow t = 3\) or \(t = 9\).
\(t = 3\): \(x^2 + 2x - 3 = 1 \Rightarrow x^2 + 2x - 4 = 0\), product of roots \(-4\). \(t = 9\): \(x^2 + 2x - 5 = 0\), product \(-5\).
Both quadratics have real roots, so the product of all four values is \((-4)(-5) = 20\).