The Correct Option is D
Solution and Explanation
Approach: Use the change-of-base shortcut $\log_{2^k}N = \frac{\log_8 N}{\log_8 2^k}$ by working entirely in base $8$ — all three given bases are powers of $2$ that relate cleanly to $8=2^3$. Get the product condition, then instead of quoting AM-GM, argue the symmetric minimum directly.
Step 1: Convert everything to powers of $2$ in one shot. Writing $a=\log_2 x,\ b=\log_2 y,\ c=\log_2 z$: \[ \log_{64}x^2 = \frac{2a}{6} = \frac{a}{3},\quad \log_8\sqrt y = \frac{b/2}{3} = \frac{b}{6},\quad 3\log_{512}(\sqrt y\,z) = 3\cdot\frac{b/2 + c}{9} = \frac{b/2+c}{3}. \]
Step 2: Sum to $4$: \[ \frac{a}{3} + \frac{b}{6} + \frac{b/2+c}{3} = 4 \;\Rightarrow\; \frac{a}{3}+\frac{b}{6}+\frac{b}{6}+\frac{c}{3} = 4 \;\Rightarrow\; \frac{a+b+c}{3} = 4. \] Hence $a+b+c = 12$, i.e. $\log_2(xyz)=12$ and $xyz = 4096$.
Step 3: Now minimise $x+y+z$ given the fixed product $xyz=4096$. For positive numbers with a fixed product, the sum is smallest when all three are equal (any imbalance — making one larger and another smaller while keeping the product fixed — increases the sum, since the function is convex). Setting $x=y=z=t$ gives $t^3 = 4096$, so $t = 16$.
Step 4: Therefore the minimum sum is \[ x+y+z = 3t = 3\times 16 = 48. \] Final answer: $48$.