<p><strong>766.</strong> Positive numbers \(x\), \(y\) and \(z\) satisfy \(xyz = 10^{81}\) and \((\log_{10} x)(\log_{10} yz) + (\log_{10} y)(\log_{10} z) = 468\). Find \(\sqrt{(\log_{10} x)^2 + (\log_{10} y)^2 + (\log_{10} z)^2}\).</p>
Step-by-Step Solution
Key Concept: Let a = log₁₀ x, b = log₁₀ y, c = log₁₀ z. Then a + b + c = 81 (from xyz = 10⁸¹) and the constraint ab + bc + ca = ab + c(a + b) = ab + c(81 - c) = 468 simplifies to finding (a² + b² + c²).
<p><strong>Step 1:</strong> Let a = log₁₀ x, b = log₁₀ y, c = log₁₀ z. From xyz = 10⁸¹, we get:</p><p>a + b + c = 81</p><p><strong>Step 2:</strong> Rewrite the constraint ab + bc + ca = 468. The given equation is (log₁₀ x)(log₁₀ yz) + (log₁₀ y)(log₁₀ z) = 468, which becomes:</p><p>a(b + c) + bc = 468</p><p>Since b + c = 81 - a:</p><p>a(81 - a) + bc = 468</p><p>81a - a² + bc = 468</p><p><strong>Step 3:</strong> Rearrange to find ab + bc + ca. Note that:</p><p>ab + bc + ca = a(b + c) + bc = a(81 - a) + bc = 468</p><p>This directly gives us: ab + ac + bc = 468</p><p><strong>Step 4:</strong> Use the identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca):</p><p>81² = a² + b² + c² + 2(468)</p><p>6561 = a² + b² + c² + 936</p><p>a² + b² + c² = 5625</p><p><strong>Step 5:</strong> Therefore:</p><p>√(a² + b² + c²) = √5625 = 75</p><p>∴ Answer: <strong>75</strong></p>
Correct Answer: 75