Q2
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
Answer: B
Key idea: Convert every term to base 2, then use AM-GM.
Since \(64 = 2^6\), \(8 = 2^3\) and \(512 = 2^9\):
- \(\log_{64} x^2 = \dfrac{2}{6}\log_2 x = \dfrac{1}{3}\log_2 x\)
- \(\log_8 \sqrt{y} = \dfrac{1}{3}\cdot\dfrac{1}{2}\log_2 y = \dfrac{1}{6}\log_2 y\)
- \(3\log_{512}(\sqrt{y}\,z) = \dfrac{3}{9}\left(\dfrac{1}{2}\log_2 y + \log_2 z\right) = \dfrac{1}{6}\log_2 y + \dfrac{1}{3}\log_2 z\)
Adding: \(\dfrac{1}{3}\left(\log_2 x + \log_2 y + \log_2 z\right) = 4\), so \(\log_2(xyz) = 12\), i.e. \(xyz = 2^{12} = 4096\).
By AM-GM, \(x+y+z \ge 3\sqrt[3]{xyz} = 3 \times 16 = 48\), with equality at \(x=y=z=16\).
Hence, the answer is (B) 48.