CAT 2025 Slot 2 QA Q2 Logarithms & Exponents

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

If \( \log _ { 64 } x ^ { 2 } + \log _ { 8 } \sqrt { y } + 3 \log _ { 512 } ( \sqrt { y } z ) = 4 \), where \( x , y \) and \( z \) are positive real numbers, then the minimum possible value of \( ( x + y + z ) \) is

A
\( 96 \)
B
\( 48 \)
C
\( 36 \)
D
\( 24 \)

Answer: (B) 48 Logarithms & Exponents MCQ

Solution

Convert to base 2: \(\log_{64}x^2 = \frac{2}{6}\log_2 x = \frac13\log_2 x\); \(\log_8\sqrt y = \frac16\log_2 y\).
\(3\log_{512}(\sqrt y\, z) = \frac{3}{9}\left(\frac12\log_2 y + \log_2 z\right) = \frac16\log_2 y + \frac13\log_2 z\).
Sum: \(\frac13(\log_2 x + \log_2 y + \log_2 z) = 4\), so \(xyz = 2^{12} = 4096\).
By AM-GM, \(x+y+z \ge 3\sqrt[3]{4096} = 3 \times 16 = 48\), with equality at \(x = y = z = 16\).

Shortcut. Once the product is fixed, the minimum sum is at equal values: \(3 \times 16\).