CAT 2025 Slot 2 · Quant · Question 2

Since 64={2}^{6}, 8={2}^{3} and 512={2}^{9}, write every term in base 2.

{\log}_{64}{x}^{2}=\frac{2{\log}_{2}x}{6}=\frac{{\log}_{2}x}{3}.

{\log}_{8}\sqrt{y}=\frac{\frac{1}{2}{\log}_{2}y}{3}=\frac{{\log}_{2}y}{6}.

3{\log}_{512}\left( \sqrt{y}\,z \right)=\frac{3\left( \frac{1}{2}{\log}_{2}y+{\log}_{2}z \right)}{9}=\frac{{\log}_{2}y}{6}+\frac{{\log}_{2}z}{3}.

Adding the three, the two \frac{{\log}_{2}y}{6} terms combine and the left side becomes \frac{1}{3}\left({\log}_{2}x+{\log}_{2}y+{\log}_{2}z \right)=4.

So {\log}_{2}\left( xyz \right)=12, giving xyz={2}^{12}=4096.

By AM-GM,
AM \ge GM

\frac{x+y+z}{3}\ge \sqrt[3]{xyz}=\sqrt[3]{4096}=16.

Hence x+y+z\ge 48, with equality at x=y=z=16, so the minimum is 48.

Shortcut. The moment all three terms reduce to sixths and thirds of {\log}_{2}, the left side has to be \frac{1}{3}{\log}_{2}\left( xyz \right) — a sum of logs of a product. So jump straight to xyz={2}^{12} and then to AM-GM; the question is a one-line AM-GM dressed up as logarithms.