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:

Notice that 64, 8 and 512 are all powers of 2 -- convert every logarithm to base 2 first so the three terms can be combined into a single equation in $\log_2 x$, $\log_2 y$ and $\log_2 z$. Once you get a fixed value for the product $xyz$, remember AM-GM gives the smallest possible sum, achieved when all three quantities are equal.

Answer 1

The Correct Option is D

Approach: Every base (\(64,8,512\)) is a power of \(2\), so convert all three logs to base \(2\). The whole left side collapses to a single symmetric expression \(\tfrac13(\log_2 x+\log_2 y+\log_2 z)\), turning the condition into a fixed product \(xyz\) — and then AM-GM gives the minimum sum.

Step 1: Write each term in base \(2\) using \(\log_{2^k}(N)=\tfrac1k\log_2 N\): \[ \log_{64}x^2 = \tfrac{1}{6}\log_2 x^2 = \tfrac{1}{3}\log_2 x, \] \[ \log_8\sqrt{y} = \tfrac{1}{3}\log_2 y^{1/2} = \tfrac{1}{6}\log_2 y, \] \[ 3\log_{512}(\sqrt{y}\,z) = 3\cdot\tfrac{1}{9}\log_2(\sqrt{y}\,z) = \tfrac{1}{3}\!\left(\tfrac12\log_2 y + \log_2 z\right). \]

Step 2: Add them and set equal to \(4\): \[ \tfrac13\log_2 x + \tfrac16\log_2 y + \tfrac16\log_2 y + \tfrac13\log_2 z = 4. \] The two \(\tfrac16\log_2 y\) terms combine to \(\tfrac13\log_2 y\), giving the clean symmetric form \[ \tfrac13\big(\log_2 x + \log_2 y + \log_2 z\big) = 4. \]

Step 3: Multiply by \(3\): \(\log_2 x+\log_2 y+\log_2 z = 12\), i.e. \[ \log_2(xyz) = 12 \;\Rightarrow\; xyz = 2^{12} = 4096. \]

Step 4: By AM-GM on positive reals, \[ \frac{x+y+z}{3} \ge \sqrt[3]{xyz} = \sqrt[3]{4096} = 16, \] so \(x+y+z \ge 48\), with equality when \(x=y=z=16\) (each is a valid positive real, and \(16\cdot16\cdot16=4096\)). The constraint imposes no extra restriction, so the bound is attainable.

Step 5: Minimum \(x+y+z = 16+16+16 = 48\). Final answer: \(48\).

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 \[ 48. \]

Answer 3

Concept:

• Every base in this equation (64, 8, 512) is a power of 2, so each logarithm can be rewritten as a base-2 logarithm of a fractional power of the variable, using $\log_{a^n}(M) = \log_a\left(M^{1/n}\right)$.
• Once every term is a base-2 log, the three terms can be merged into a single logarithm of one combined product using $\log_a M + \log_a N = \log_a(MN)$, instead of adding separate fractional coefficients.
• For positive numbers with a fixed product, AM-GM gives the smallest possible sum, with equality when all the numbers are equal.

Step 1: Rewrite each logarithm as a base-2 log of a power of the variable.
$\log_{64}x^2 = \log_{2^6}x^2 = \log_2\left(x^{2/6}\right) = \log_2\left(x^{1/3}\right)$
$\log_8\sqrt{y} = \log_{2^3}y^{1/2} = \log_2\left(y^{1/6}\right)$
$3\log_{512}(\sqrt{y}\,z) = 3\log_{2^9}(\sqrt{y}\,z) = \log_2\left((\sqrt{y}\,z)^{1/3}\right) = \log_2\left(y^{1/6}z^{1/3}\right)$

Step 2: Merge all three logs into one, using the product rule.
$\log_2\left(x^{1/3}\right) + \log_2\left(y^{1/6}\right) + \log_2\left(y^{1/6}z^{1/3}\right) = \log_2\left(x^{1/3}y^{1/6}y^{1/6}z^{1/3}\right) = \log_2\left(x^{1/3}y^{1/3}z^{1/3}\right)$
Since $x^{1/3}y^{1/3}z^{1/3} = (xyz)^{1/3}$, the equation becomes $\log_2\left((xyz)^{1/3}\right) = 4$.

Step 3: Solve for the product $xyz$.
$(xyz)^{1/3} = 2^4 = 16$, so $xyz = 16^3 = 4096$.

Step 4: Minimize $x+y+z$ for positive reals with a fixed product, using AM-GM.
$\dfrac{x+y+z}{3} \ge \sqrt[3]{xyz} = \sqrt[3]{4096} = 16$, so $x+y+z \ge 48$, with equality when $x=y=z=16$.

Final Answer: $48$