Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If 1, \(\log_9(3^{1-x} + 2)\), \(\log_3(4 \cdot 3^x - 1)\) are in A.P., then \(x\) equals</p>
<p>(A) \(\log_3 4\)</p>
<p>(B) \(1 - \log_3 4\)</p>
<p>(C) \(1 - \log_4 3\)</p>
<p>(D) \(\log_4 3\)</p>
Step-by-Step Solution
Key Concept: Use the A.P. condition that the middle term equals the average of the first and third terms, and convert logarithms to the same base.
<p><strong>Step 1:</strong> Since 1, \(\log_9(3^{1-x} + 2)\), \(\log_3(4 \cdot 3^x - 1)\) are in A.P., we use the condition: \(2 \cdot \log_9(3^{1-x} + 2) = 1 + \log_3(4 \cdot 3^x - 1)\)</p><p><strong>Step 2:</strong> Convert \(\log_9\) to base 3: \(\log_9(3^{1-x} + 2) = \frac{1}{2}\log_3(3^{1-x} + 2)\)</p><p><strong>Step 3:</strong> Substitute: \(\log_3(3^{1-x} + 2) = 1 + \log_3(4 \cdot 3^x - 1)\)</p><p><strong>Step 4:</strong> This gives: \(3^{1-x} + 2 = 3(4 \cdot 3^x - 1)\)</p><p><strong>Step 5:</strong> Simplify: \(3 \cdot 3^{-x} + 2 = 12 \cdot 3^x - 3\)</p><p><strong>Step 6:</strong> Let \(y = 3^x\): \(3/y + 2 = 12y - 3\), which gives \(12y^2 - 5y - 3 = 0\)</p><p><strong>Step 7:</strong> Solving: \(y = 3^x = \frac{4}{3}\), so \(x = 1 - \log_3 4\)</p><p>∴ Answer is B.</p>
Correct Answer: B