<p>If \(1,\ \log_{\sqrt{3^{1-x}+2}},\ \log_3(4\cdot3^x - 1)\) are in AP, then \(x\) equals</p>
Step-by-Step Solution
Key Concept: If three terms are in AP, then the middle term equals the average of the first and third terms. Use 2b = a + c, then convert logarithmic equation to exponential form by recognizing that log₃(base)² = 2log₃(base).
<p><strong>Step 1:</strong> Apply AP condition: If 1, log√(3^(1-x)+2), log₃(4·3^x - 1) are in AP, then:</p><p>2·log√(3^(1-x)+2) = 1 + log₃(4·3^x - 1)</p><p><strong>Step 2:</strong> Convert log√ to base 3. Since √(3^(1-x)+2) = (3^(1-x)+2)^(1/2), we have:</p><p>log√(3^(1-x)+2) = (1/2)log₃(3^(1-x)+2)</p><p>So: 2·(1/2)log₃(3^(1-x)+2) = 1 + log₃(4·3^x - 1)</p><p>log₃(3^(1-x)+2) = 1 + log₃(4·3^x - 1)</p><p><strong>Step 3:</strong> Rewrite: log₃(3^(1-x)+2) = log₃3 + log₃(4·3^x - 1)</p><p>log₃(3^(1-x)+2) = log₃[3(4·3^x - 1)]</p><p><strong>Step 4:</strong> Equate arguments:</p><p>3^(1-x) + 2 = 3(4·3^x - 1)</p><p>3·3^(-x) + 2 = 12·3^x - 3</p><p>3·3^(-x) = 12·3^x - 5</p><p><strong>Step 5:</strong> Let y = 3^x. Then 3^(-x) = 1/y:</p><p>3/y = 12y - 5</p><p>3 = 12y² - 5y</p><p>12y² - 5y - 3 = 0</p><p>(3y + 1)(4y - 3) = 0</p><p><strong>Step 6:</strong> Since 3^x > 0, we have y = 3/4, so 3^x = 3/4</p><p>x = log₃(3/4) = log₃3 - log₃4 = 1 - log₃4</p><p>Or: x = log₃(3/4) = -log₃(4/3)</p><p>∴ Answer: B</p>
Correct Answer: B