<p>If \(\alpha = 2\beta = 3\gamma\), then find the value of \(\log_{\left(\frac{7}{6}\right)}\left(\dfrac{\alpha^2 + \beta^2 + \gamma^2}{\alpha\beta + \beta\gamma + \gamma\alpha}\right)\).</p>
Step-by-Step Solution
Key Concept: Express all variables in terms of a single parameter using the constraint α = 2β = 3γ, then compute the numerator and denominator as quadratic expressions to find their ratio, which becomes an argument for the logarithm.
<p><strong>Step 1:</strong> Use the constraint α = 2β = 3γ to express everything in terms of one variable.</p><p>Let α = 2β = 3γ = 6k for some k ≠ 0.</p><p>Then: α = 6k, β = 3k, γ = 2k</p><p><strong>Step 2:</strong> Calculate the numerator α² + β² + γ².</p><p>α² + β² + γ² = (6k)² + (3k)² + (2k)² = 36k² + 9k² + 4k² = 49k²</p><p><strong>Step 3:</strong> Calculate the denominator αβ + βγ + γα.</p><p>αβ + βγ + γα = (6k)(3k) + (3k)(2k) + (2k)(6k) = 18k² + 6k² + 12k² = 36k²</p><p><strong>Step 4:</strong> Find the ratio inside the logarithm.</p><p>$$\dfrac{\alpha^2 + \beta^2 + \gamma^2}{\alpha\beta + \beta\gamma + \gamma\alpha} = \dfrac{49k^2}{36k^2} = \dfrac{49}{36} = \left(\dfrac{7}{6}\right)^2$$</p><p><strong>Step 5:</strong> Evaluate the logarithm.</p><p>$$\log_{\left(\frac{7}{6}\right)}\left(\left(\dfrac{7}{6}\right)^2\right) = 2$$</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2