CAT 2023 Slot 2 QA Q2 Logarithms & Exponents

Question 2 of the Quantitative Ability (QA) section, CAT 2023 Slot 2, with CAT's official answer and a worked solution.

The sum of all possible values of \(x\) satisfying the equation \(2^{4 x^2}-2^{2 x^2+x+16}+2^{2 x+30}=0\), is
  1. A
    \(\frac{3}{2}\)
  2. B
    \(\frac{5}{2}\)
  3. C
    \(\frac{1}{2}\)
  4. D
    3

Answer: (C) 1/2 Logarithms & Exponents MCQ

Solution

  1. Let \(u = 2^{2x^2}\) and \(v = 2^{x+15}\). Then \(2^{4x^2} = u^2\), \(2^{2x^2+x+16} = 2uv\) and \(2^{2x+30} = v^2\).
  2. The equation becomes \(u^2 - 2uv + v^2 = (u - v)^2 = 0\), so \(u = v\).
  3. \(2x^2 = x + 15 \Rightarrow 2x^2 - x - 15 = 0\), i.e. \((2x+5)(x-3)=0\), so \(x = 3\) or \(x = -\frac{5}{2}\).
  4. Sum of the roots \(= 3 - \frac{5}{2} = \frac{1}{2}\).

Shortcut. Spot the perfect square \((u-v)^2\); the sum of roots of \(2x^2 - x - 15\) is \(\frac{1}{2}\).

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