Basic Mathematics & Logarithm
Logarithmic simplification
Grade 11
Question:
<p>Let \(a = 7^{\dfrac{1}{\log_8 \sqrt{343}}}\) and \(b = 11^{\dfrac{1}{\log_5 \sqrt{11}}}\), then:</p>
<p>both \(a\) and \(b\) are prime</p>
<p>\(a\) and \(b\) are relatively prime</p>
<p>both \(a\) and \(b\) are irrational</p>
<p>\(a\) and \(b\) both are rational but not integer</p>
Step-by-Step Solution
Key Concept: Use the change of base formula: 1/logₐb = logᵦa. This converts the reciprocal logarithm in the exponent into a standard logarithm, allowing direct simplification.
<p><strong>Step 1:</strong> Apply the change of base identity 1/logₐb = logᵦa</p><p>For a: Note that 1/log₈(√343) = log_{√343}(8)</p><p>Therefore: a = 7^{log_{√343}(8)}</p><p><strong>Step 2:</strong> Simplify √343 = √(7³) = 7^(3/2)</p><p>So a = 7^{log_{7^(3/2)}(8)}</p><p><strong>Step 3:</strong> Use the property: if base = 7^k, then log_{7^k}(x) = (1/k)log₇(x)</p><p>Thus: log_{7^(3/2)}(8) = (2/3)log₇(8)</p><p>Therefore: a = 7^{(2/3)log₇(8)} = (7^{log₇(8)})^{2/3} = 8^{2/3}</p><p><strong>Step 4:</strong> Similarly for b: 1/log₅(√11) = log_{√11}(5)</p><p>Since √11 = 11^(1/2): log_{11^(1/2)}(5) = 2log₁₁(5)</p><p>Therefore: b = 11^{2log₁₁(5)} = (11^{log₁₁(5)})² = 5² = 25</p><p><strong>Step 5:</strong> Calculate a = 8^{2/3} = (2³)^{2/3} = 2² = 4</p><p>∴ a = 4 and b = 25</p>
Correct Answer: A