CAT 2023 Slot 1 QA Q1 Logarithms & Exponents

Question 1 of the Quantitative Ability (QA) section, CAT 2023 Slot 1, with CAT's official answer and a worked solution.

If \(x\) and \(y\) are positive real numbers such that \(\log _x\left(x^2+12\right)=4\) and \(3 \log _y x=1\), then \(x+y\) equals
  1. A
    20
  2. B
    11
  3. C
    68
  4. D
    10

Answer: (D) 10 Logarithms & Exponents MCQ

Solution

  1. \(\log_x(x^2+12)=4\) means \(x^4 = x^2 + 12\), i.e. \((x^2-4)(x^2+3)=0\), so \(x^2=4\) and \(x=2\).
  2. \(3\log_y x = 1\) gives \(\log_y 2 = \frac{1}{3}\), so \(y = 2^3 = 8\).
  3. \(x + y = 2 + 8 = 10\).

Shortcut. Try \(x=2\): \(\log_2 16 = 4\) works immediately.

More Logarithms & Exponents questions from CAT

All 55 Logarithms & Exponents questions asked since 2017.