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

To solve the problem, we need to use logarithmic properties to simplify the given equation:

The equation given is:

\(\log_{64} x^2 + \log_8 \sqrt{y} + 3 \log_{512} (\sqrt{y} z) = 4\)

We will rewrite the logs in terms of base 2, since all bases 64, 8, and 512 are powers of 2:

1. \(64 = 2^6\), so \(\log_{64} x^2 = \frac{1}{6} \log_{2} x^2 = \frac{2}{6} \log_{2} x = \frac{1}{3} \log_{2} x\)
2. \(8 = 2^3\), so \(\log_8 \sqrt{y} = \frac{1}{3} \log_{2} y^{1/2} = \frac{1}{6} \log_{2} y\)
3. \(512 = 2^9\), so \(3 \log_{512} (\sqrt{y} z) = 3 \times \frac{1}{9} \log_{2} (\sqrt{y} z) = \frac{1}{3} \log_{2} (\sqrt{y} z)\)

Substitute these into the original equation:

\(\frac{1}{3} \log_{2} x + \frac{1}{6} \log_{2} y + \frac{1}{3} \log_{2} (\sqrt{y} z) = 4\)

Simplify further:

The term \(\log_{2} (\sqrt{y} z)\) becomes \(\log_{2} y^{1/2} + \log_{2} z = \frac{1}{2} \log_{2} y + \log_{2} z\). Substitute this back:

\(\frac{1}{3} \log_{2} x + \frac{1}{6} \log_{2} y + \frac{1}{3} \left(\frac{1}{2} \log_{2} y + \log_{2} z\right) = 4\)

Combine the terms:

\(\frac{1}{3} \log_{2} x + \frac{1}{6} \log_{2} y + \frac{1}{6} \log_{2} y + \frac{1}{3} \log_{2} z = 4\)

This simplifies to:

\(\frac{1}{3} \log_{2} x + \frac{1}{3} \log_{2} y + \frac{1}{3} \log_{2} z = 4\)

Now, multiply through by 3 to clear fractions:

\(\log_{2} x + \log_{2} y + \log_{2} z = 12\)

Using the property \(\log_{2} a + \log_{2} b + \log_{2} c = \log_{2} (abc)\), we have:

\(\log_{2} (xyz) = 12\)

Thus, \(xyz = 2^{12} = 4096\).

To find the minimum of \(x+y+z\) subject to \(xyz = 4096\), use the AM-GM inequality which states:

\(\frac{x+y+z}{3} \geq \sqrt[3]{xyz}\)

Thus:

\(\frac{x+y+z}{3} \geq \sqrt[3]{4096} = \sqrt[3]{2^{12}} = 8\)

Therefore, \(x+y+z \geq 24\).

However, calculating directly for minimum values, we find equality in AM-GM when \(x = y = z\):

Substitute \(x = y = z = k\) gives \(k^3 = 4096\), so \(k = \sqrt[3]{4096} = 16\).

Thus, \(x+y+z = 16 + 16 + 16 = 48\).

Therefore, the minimum possible value of \(x+y+z\) is 48.

Answer 2

Step 1: Rewrite all logarithms with base \(2\). Since \(64 = 2^6\), \(8 = 2^3\), and \(512 = 2^9\), we convert each logarithmic term: \[ \log_{64} x^2 = \log_{2^6} x^2 = \frac{\log_2 x^2}{\log_2 2^6} = \frac{2\log_2 x}{6} = \frac{1}{3}\log_2 x. \] \[ \log_8 \sqrt{y} = \log_{2^3} y^{1/2} = \frac{\log_2 y^{1/2}}{\log_2 2^3} = \frac{\frac{1}{2}\log_2 y}{3} = \frac{1}{6}\log_2 y. \] \[ 3\log_{512}(\sqrt{y}z) = 3\log_{2^9}(y^{1/2}z) = 3\cdot \frac{\log_2(y^{1/2}z)}{9} = \frac{1}{3}\left(\frac{1}{2}\log_2 y + \log_2 z\right) = \frac{1}{6}\log_2 y + \frac{1}{3}\log_2 z. \] Step 2: Substitute into the given equation. The equation \[ \log_{64} x^2 + \log_8 \sqrt{y} + 3\log_{512}(\sqrt{y}z) = 4 \] becomes \[ \frac{1}{3}\log_2 x + \frac{1}{6}\log_2 y + \left(\frac{1}{6}\log_2 y + \frac{1}{3}\log_2 z\right) = 4. \] Combining like terms, \[ \frac{1}{3}\log_2 x + \frac{1}{3}\log_2 y + \frac{1}{3}\log_2 z = 4. \] Factoring \(\frac{1}{3}\), \[ \frac{1}{3}\bigl(\log_2 x + \log_2 y + \log_2 z\bigr) = 4. \] Using the logarithmic identity, \[ \log_2 x + \log_2 y + \log_2 z = \log_2(xyz), \] we obtain \[ \frac{1}{3}\log_2(xyz) = 4 \Rightarrow \log_2(xyz) = 12. \] Thus, \[ xyz = 2^{12} = 4096. \] Step 3: Minimize \(x + y + z\) using AM–GM inequality. For positive real numbers \(x, y, z\), \[ \frac{x + y + z}{3} \ge \sqrt[3]{xyz}. \] Substituting \(xyz = 2^{12}\), \[ \frac{x + y + z}{3} \ge \sqrt[3]{2^{12}} = 2^{4} = 16. \] Hence, \[ x + y + z \ge 48. \] Equality holds when \(x = y = z\), so the minimum value of \(x + y + z\) is achieved at \[ x = y = z = 16. \] Therefore, the minimum possible value of \(x + y + z\) is \[ 48. \]