If $\log_{64} x^2 + \log_8 \sqrt{y} + 3\log_{512}\left(\sqrt{y}z\right) = 4$, where $x, y$ and $z$ are positive real numbers, then the minimum possible value of $(x+y+z)$ is:

Answer: 48

Using $\log_{2^k} a = \frac{1}{k}\log_2 a$, and writing $\log$ for $\log_2$:

• $\log_{64} x^2 = \frac{2\log x}{6} = \frac{1}{3}\log x$
• $\log_8 \sqrt{y} = \frac{\frac{1}{2}\log y}{3} = \frac{1}{6}\log y$
• $3\log_{512}\left(\sqrt{y}\,z\right) = 3 \cdot \frac{\frac{1}{2}\log y + \log z}{9} = \frac{1}{6}\log y + \frac{1}{3}\log z$

Simplifying the equation:

$\frac{1}{3}\log x + \frac{1}{6}\log y + \frac{1}{6}\log y + \frac{1}{3}\log z = 4$

$\frac{1}{3}\left(\log x + \log y + \log z\right) = 4 \implies \log(xyz) = 12 \implies xyz = 2^{12}$

Now apply AM-GM

$x + y + z \ge 3\sqrt[3]{xyz} = 3\sqrt[3]{2^{12}} = 3 \cdot 2^4 = 48$

Equality holds when $x = y = z = 16$.

Check: $xyz = 16^3 = 2^{12}$, which satisfies the constraint.

Minimum value of $(x + y + z)$ = 48