<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 given quantities a and b by breaking down 25, 225, 81, and 2250 into prime factors, then use logarithm properties to combine terms.
<p><strong>Step 1:</strong> Express a and b in terms of prime factors:</p><p>• a = log 25 = log(5²) = 2log 5</p><p>• b = log 225 = log(9 × 25) = log(3²) + log(5²) = 2log 3 + 2log 5 = 2log 3 + a</p><p>∴ log 3 = (b - a)/2</p><p><strong>Step 2:</strong> Decompose the expression using logarithm properties:</p><p>log(1/81) + log(1/2250) = -log 81 - log 2250</p><p>= -log(3⁴) - log(2 × 1125) = -log(3⁴) - log(2 × 9 × 125)</p><p>= -4log 3 - log 2 - 2log 3 - 3log 5</p><p>= -6log 3 - log 2 - 3log 5</p><p><strong>Step 3:</strong> Express in terms of a and b:</p><p>Since a = 2log 5, we have log 5 = a/2</p><p>Since log 3 = (b - a)/2</p><p>And log 2 = log(10/5) = log 10 - log 5 = 1 - a/2</p><p><strong>Step 4:</strong> Substitute:</p><p>-6log 3 - log 2 - 3log 5 = -6·(b-a)/2 - (1 - a/2) - 3·(a/2)</p><p>= -3(b - a) - 1 + a/2 - 3a/2</p><p>= -3b + 3a - 1 - a</p><p>= -3b + 2a - 1</p><p>∴ Answer: C</p>
Correct Answer: C