Minimum Value of Logarithmic Expression

Simplify each log base to base 2:

• $\log_{64} x^2 = \frac{2}{6} \log_2 x = \frac{1}{3} \log_2 x$.
• $\log_8 \sqrt{y} = \frac{1/2}{3} \log_2 y = \frac{1}{6} \log_2 y$.
• $3 \log_{512} (\sqrt{y} z) = 3 \cdot \frac{1}{9} \log_2 (y^{1/2} z) = \frac{1}{3} (\frac{1}{2} \log_2 y + \log_2 z) = \frac{1}{6} \log_2 y + \frac{1}{3} \log_2 z$. Sum = $\frac{1}{3} \log_2 x + \frac{1}{3} \log_2 y + \frac{1}{3} \log_2 z = 4$. $\frac{1}{3} \log_2 (xyz) = 4 \implies \log_2 (xyz) = 12 \implies xyz = 2^{12} = 4096$. To minimize $x + y + z$ (noting expression $x+y+z$), by AM-GM, $x = y = z = (4096)^{1/3} = 16$. Minimum sum $x + y + z = 16 + 16 + 16 = 48$. Wait, if expression is $x + y + x = 2x + y$, or $x + y + z = 48$, option 4 is 48.