<p><strong>178.</strong> Let \(a = \log 25\) and \(b = \log 225\), then \(\log\!\left(\dfrac{1}{81}\right) + \log\!\left(\dfrac{1}{2250}\right)\) is equal to:</p>
Step-by-Step Solution
Key Concept: Express all logarithms in terms of basic logarithmic values (log 2, log 3, log 5) by decomposing numbers into prime factors, then use logarithm properties to combine terms.
<p><strong>Step 1:</strong> Express the given logarithms using prime factorization.</p><p>• a = log 25 = log 5² = 2 log 5</p><p>• b = log 225 = log (9 × 25) = log 3² + log 5² = 2 log 3 + 2 log 5</p><p><strong>Step 2:</strong> Simplify the target expression using logarithm properties.</p><p>log(1/81) + log(1/2250) = -log 81 - log 2250</p><p>= -log 3⁴ - log(9 × 250)</p><p>= -4 log 3 - log 3² - log (2 × 125)</p><p>= -4 log 3 - 2 log 3 - log 2 - log 5³</p><p>= -6 log 3 - log 2 - 3 log 5</p><p><strong>Step 3:</strong> Recognize that without additional context (like assuming base 10 where specific values apply), express in standard form or identify the numerical answer using standard logarithm tables if needed.</p><p>The expression simplifies to: <strong>-log(2 × 3⁶ × 5³)</strong> or equivalently <strong>-log(2 × 729 × 125) = -log 91,125</strong></p><p>∴ Answer: C</p>
Correct Answer: C