Question

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

• A. \(24\)
• B. \(36\)
• C. \(96\)
• D. \(48\)

Hint:

When an equation in logarithms simplifies to a fixed product like $xyz = \text{constant}$, use the AM-GM inequality to find the minimum (or maximum) of sums such as $x + y + z$. Equality in AM-GM occurs when all the variables are equal.

Answer 1

The Correct Option is D

Step 1: Convert all logarithms to base $2$. Since $64 = 2^6$, $8 = 2^3$, and $512 = 2^9$: \[ \log_{64} x^2 = \log_{2^6} x^2 = \frac{\log_2 x^2}{\log_2 2^6} = \frac{2 \log_2 x}{6} = \frac{1}{3} \log_2 x. \] \[ \log_8 \sqrt{y} = \log_{2^3} y^{1/2} = \frac{\log_2 y^{1/2}}{\log_2 2^3} = \frac{\frac{1}{2}\log_2 y}{3} = \frac{1}{6} \log_2 y. \] \[ 3\log_{512}(\sqrt{y} z) = 3\log_{2^9}(y^{1/2} z) = 3\cdot \frac{\log_2(y^{1/2} z)}{\log_2 2^9} = 3\cdot \frac{\frac{1}{2}\log_2 y + \log_2 z}{9} = \frac{1}{3}\left(\frac{1}{2}\log_2 y + \log_2 z\right) = \frac{1}{6}\log_2 y + \frac{1}{3}\log_2 z. \]
Step 2: Substitute into the given equation. \[ \log_{64} x^2 + \log_{8} \sqrt{y} + 3\log_{512}(\sqrt{y}z) = 4 \] becomes \[ \frac{1}{3}\log_2 x + \frac{1}{6}\log_2 y + \left( \frac{1}{6}\log_2 y + \frac{1}{3}\log_2 z \right) = 4. \] Combine like terms: \[ \frac{1}{3}\log_2 x + \left(\frac{1}{6} + \frac{1}{6}\right)\log_2 y + \frac{1}{3}\log_2 z = 4 \] \[ \frac{1}{3}\log_2 x + \frac{1}{3}\log_2 y + \frac{1}{3}\log_2 z = 4. \] Factor $\dfrac{1}{3}$: \[ \frac{1}{3}\bigl(\log_2 x + \log_2 y + \log_2 z\bigr) = 4. \] Using $\log_2 x + \log_2 y + \log_2 z = \log_2(xyz)$: \[ \frac{1}{3}\log_2(xyz) = 4 \;\Rightarrow\; \log_2(xyz) = 12. \] Therefore, \[ xyz = 2^{12} = 4096. \]
Step 3: Minimize $x + y + z$ using AM-GM. For positive $x,y,z$, \[ \frac{x + y + z}{3} \ge \sqrt[3]{xyz}. \] Since $xyz = 2^{12}$: \[ \frac{x + y + z}{3} \ge \sqrt[3]{2^{12}} = 2^{12/3} = 2^4 = 16. \] Thus, \[ x + y + z \ge 3 \cdot 16 = 48. \] The equality in AM-GM holds when $x = y = z$, so the minimum sum $x + y + z$ is $48$, attained when \[ x = y = z = 16. \] Therefore, the minimum possible value of $(x + y + z)$ is \(48\).