⚡ AZUCATION SHORTCUT:
Simplify to base 8: $\log_8 x + \log_8 \sqrt{y} + \log_8 \sqrt{yz} = 4 \implies xy\sqrt{z} = 8^4 = 4096$.
For min sum, $x = y = 2z$.
$(2z)(2z)\sqrt{z} = 4096 \implies 4z^{2.5} = 4096 \implies z = 16$.
Sum $= 32 + 32 + 16 = \mathbf{80}$.

Step 1: Simplify Logarithmic Terms
Convert all terms to base 8:
1. $\log_{64} x^2 = \log_{8^2} x^2 = \frac{2}{2} \log_8 x = \log_8 x$
2. $3 \log_{512} \sqrt{yz} = 3 \log_{8^3} (yz)^{1/2} = 3 \cdot \frac{1}{3} \log_8 (yz)^{1/2} = \log_8 \sqrt{yz}$

Step 2: Combine the Equation
$\log_8 x + \log_8 \sqrt{y} + \log_8 \sqrt{yz} = 4$
$\log_8 (x \cdot \sqrt{y} \cdot \sqrt{y} \cdot \sqrt{z}) = 4$
$\log_8 (x \cdot y \cdot \sqrt{z}) = 4 \implies xy\sqrt{z} = 8^4 = 4096$

Step 3: Apply AM-GM for Minimum Value
To minimize $x+y+z$ given $x^1 y^1 z^{0.5} = 4096$, the values must be proportional to their exponents:
$\frac{x}{1} = \frac{y}{1} = \frac{z}{0.5} = k \implies x=k, y=k, z=0.5k$
Substitute in product: $k \cdot k \cdot \sqrt{0.5k} = 4096 \implies k^2 \cdot \frac{\sqrt{k}}{\sqrt{2}} = 4096$
$k^{2.5} = 4096 \times \sqrt{2} = 2^{12} \cdot 2^{0.5} = 2^{12.5}$
$k = (2^{12.5})^{1/2.5} = 2^5 = 32$.
Values: $x=32, y=32, z=16$.
Minimum Sum: $32 + 32 + 16 = \mathbf{80}$.

WhatsApp Telegram 🔗 Copy Link