Basic Mathematics & Logarithm
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 get relationships between α, β, and γ, then use the third constraint to find γ and verify which statements are consistent with all three conditions simultaneously.
<p><strong>Step 1: Convert to exponential form</strong></p><p>From log_γ(2α) = 1/3: 2α = γ^(1/3)</p><p>From log_γ(5β) = 1/6: 5β = γ^(1/6)</p><p>From log_γ(αβ) = 3/2: αβ = γ^(3/2)</p><p><strong>Step 2: Express α and β in terms of γ</strong></p><p>α = γ^(1/3)/2</p><p>β = γ^(1/6)/5</p><p><strong>Step 3: Use the third constraint</strong></p><p>αβ = (γ^(1/3)/2)(γ^(1/6)/5) = γ^(1/3 + 1/6)/10 = γ^(1/2)/10</p><p>But we also have αβ = γ^(3/2)</p><p>Therefore: γ^(1/2)/10 = γ^(3/2)</p><p>γ^(1/2) = 10·γ^(3/2)</p><p>1 = 10·γ</p><p>γ = 1/10</p><p><strong>Step 4: Calculate α and β</strong></p><p>α = (1/10)^(1/3)/2 = 1/(2·10^(1/3))</p><p>β = (1/10)^(1/6)/5 = 1/(5·10^(1/6))</p><p><strong>Step 5: Verify consistency</strong></p><p>αβ = 1/(10·10^(1/2)) = 1/(10^(3/2)) = (1/10)^(3/2) ✓</p><p>∴ Answer: ACD</p>
Correct Answer: ACD