Basic Mathematics & Logarithm
System of logarithmic equations
Grade 11
Question:
<p>Let \(\alpha, \beta, \gamma\) are positive real numbers such that \(\log_\gamma(2\alpha) = \dfrac{1}{3}\), \(\log_\gamma(5\beta) = \dfrac{1}{6}\) and \(\log_\gamma(\alpha\beta) = \dfrac{3}{2}\), then:</p>
<p>(a) \(\alpha^3 = \dfrac{1}{16}\)</p>
<p>(b) \(\alpha^3 = \dfrac{1}{80}\)</p>
<p>(c) \(\gamma = \dfrac{1}{10}\)</p>
<p>(d) \(\beta^{12} = \dfrac{5^{14}}{2}\)</p>
Step-by-Step Solution
Key Concept: Convert logarithmic equations to exponential form to establish relationships between α, β, and γ, then use the constraint equation to find γ and verify which statements are consistent with all three conditions simultaneously.
<p><strong>Step 1: Convert logarithmic equations to exponential form</strong></p><p>From log_γ(2α) = 1/3: 2α = γ^(1/3) → α = γ^(1/3)/2</p><p>From log_γ(5β) = 1/6: 5β = γ^(1/6) → β = γ^(1/6)/5</p><p><strong>Step 2: Use the third constraint</strong></p><p>From log_γ(αβ) = 3/2: αβ = γ^(3/2)</p><p>Substituting: (γ^(1/3)/2)(γ^(1/6)/5) = γ^(3/2)</p><p>γ^(1/3 + 1/6)/10 = γ^(3/2)</p><p>γ^(1/2)/10 = γ^(3/2)</p><p>1/10 = γ</p><p><strong>Step 3: Find α and β</strong></p><p>α = (1/10)^(1/3)/2 = 1/(2·∛10)</p><p>β = (1/10)^(1/6)/5 = 1/(5·√(6)√10)</p><p><strong>Step 4: Verify relationships</strong></p><p>• 2α = 1/∛10 ✓</p><p>• 5β = 1/√(6)√10 ✓</p><p>• αβ = 1/(10·∛10·√(6)√10) = (1/10)^(3/2) ✓</p><p>• γ = 1/10 (positive) ✓</p><p>∴ Answer: ACD</p>
Correct Answer: ACD