NinetynineCAT 2023 Slot 3 QA Q1 Logarithms & Exponents
Question 1 of the Quantitative Ability (QA) section, CAT 2023 Slot 3, with CAT's official answer and a worked solution.
For a real number \(x\), if \(\frac{1}{2}, \frac{\log _3\left(2^x-9\right)}{\log _3 4}\), and \(\frac{\log _5\left(2^x+\frac{17}{2}\right)}{\log _5 4}\) are in an arithmetic progression, then the common difference is
- A\(\log _4 7\)
- B\(\log _4\left(\frac{3}{2}\right)\)
- C\(\log _4\left(\frac{7}{2}\right)\)
- D\(\log _4\left(\frac{23}{2}\right)\)
Answer: (C) log _4(7/2) Logarithms & Exponents MCQ
Solution
- By change of base the terms are \(\frac12,\ \log_4(2^x-9),\ \log_4(2^x+\frac{17}{2})\). Let \(t = 2^x\).
- AP condition: \(2\log_4(t-9) = \frac12 + \log_4(t+\frac{17}{2})\), i.e. \((t-9)^2 = 2(t+\frac{17}{2}) = 2t+17\).
- So \(t^2-20t+64=0\), giving \(t=16\) or \(t=4\). Since \(t-9>0\), \(t=16\).
- Terms: \(\frac12 = \log_4 2,\ \log_4 7,\ \log_4 24.5\). Common difference \(= \log_4 7 - \log_4 2 = \log_4\frac{7}{2}\).
Shortcut. Write \(\frac12\) as \(\log_4 2\) so the AP becomes a GP inside the logs: 2, 7, 24.5.