CAT 2024 Slot 2 QA Q9 Logarithms & Exponents

Question 9 of the Quantitative Ability (QA) section, CAT 2024 Slot 2, with CAT's official answer and a worked solution.

If \(a, b\) and \(c\) are positive real numbers such that \(a>10 \geq b \geq c\) and \(\frac{\log _8(a+b)}{\log _2 c}+\frac{\log _{27}(a-b)}{\log _3 c}=\frac{2}{3}\), then the greatest possible integer value of \(a\) is<

TITA: type in the answer. There are no options and no negative marking.

Answer: 14 Logarithms & Exponents TITA

Solution

\(\frac{\log_8(a+b)}{\log_2 c} = \frac{\frac13\log_2(a+b)}{\log_2 c} = \frac13\log_c(a+b)\). Similarly the second term is \(\frac13\log_c(a-b)\).
So \(\frac13\log_c(a^2-b^2) = \frac23\), giving \(a^2 - b^2 = c^2\), i.e. \(a^2 = b^2 + c^2\).
With \(c \le b \le 10\), \(a^2 \le 100 + 100 = 200\), so \(a \le \sqrt{200} \approx 14.14\).
\(a = 14\) works: \(b = 10\), \(c = \sqrt{96} \approx 9.8\) (\(c \le b\), \(c \ne 1\)). Greatest integer \(a\) is 14.