CAT 2025

Slot 2

QA

Question & Solution

1. Concept Used

• Topic: Logarithms (Change of Base) + AM-GM Inequality
• Formula: $$\log_{b^m} a^n = \frac{n}{m} \log_b a \quad \text{and} \quad \frac{x + y + z}{3} \geq \sqrt[3]{xyz}$$

2. Calculation

We are given: $$\log_{64} x^2 + \log_{8} \sqrt{y} + 3\log_{512}(\sqrt{y}z) = 4$$

Note that $64 = 8^2$ and $512 = 8^3$. We express all logarithms in base $8$ using the power rule $\log_{b^m} a^n = \frac{n}{m} \log_b a$.

Term 1: $$\log_{64} x^2 = \log_{8^2} x^2 = \frac{2}{2} \log_8 x = \log_8 x$$

Term 2: $$\log_8 \sqrt{y} = \log_8 y^{1/2} = \frac{1}{2} \log_8 y$$

Term 3: $$3\log_{512}(\sqrt{y}z) = 3 \cdot \log_{8^3}(\sqrt{y}z) = 3 \cdot \frac{1}{3} \log_8(\sqrt{y}z) = \log_8(\sqrt{y}z) = \frac{1}{2}\log_8 y + \log_8 z$$

Substituting all three terms back: $$\log_8 x + \frac{1}{2}\log_8 y + \frac{1}{2}\log_8 y + \log_8 z = 4$$

$$\log_8 x + \log_8 y + \log_8 z = 4$$

$$\log_8 (xyz) = 4$$

$$xyz = 8^4 = (2^3)^4 = 2^{12} = 4096$$

Now applying the AM-GM Inequality to find the minimum value of $x + y + z$:

$$\frac{x + y + z}{3} \geq \sqrt[3]{xyz} = \sqrt[3]{2^{12}} = 2^4 = 16$$

$$x + y + z \geq 3 \times 16 = 48$$

The minimum is achieved when $x = y = z = 16$, which satisfies $xyz = 16^3 = 4096 = 2^{12}$. ✓

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3. Solution

Answer = Option A ✅

The minimum possible value of $(x + y + z)$ is $\mathbf{48}$.