<p>The value of \(x\), for which the 6th term in the expansion of \(\left[2^{\log_2\sqrt{(9^{x-1}+7)}} + \dfrac{1}{2^{\frac{1}{5}\log_2(3^{x-1}+1)}}\right]^7\) is 84, is equal to</p>
Step-by-Step Solution
Key Concept: Simplify the logarithmic expressions in the binomial base first: 2^(log₂√(9^(x-1)+7)) = √(9^(x-1)+7) and the second term simplifies using logarithm properties, then apply the binomial theorem with T₆ = C(7,5)a²b⁵ = 84.
<p><strong>Step 1: Simplify the base expressions</strong></p><p>Let a = 2^(log₂√(9^(x-1)+7)) = √(9^(x-1)+7)</p><p>Let b = 1/2^(⅕log₂(3^(x-1)+1)) = 1/(3^(x-1)+1)^(1/5) = (3^(x-1)+1)^(-1/5)</p><p><strong>Step 2: Identify the 6th term</strong></p><p>In expansion of (a+b)⁷, the 6th term is T₆ = T_{5+1} = C(7,5)·a²·b⁵</p><p>T₆ = 21·[√(9^(x-1)+7)]²·[(3^(x-1)+1)^(-1/5)]⁵</p><p>T₆ = 21·(9^(x-1)+7)·(3^(x-1)+1)^(-1)</p><p><strong>Step 3: Set up the equation</strong></p><p>21·(9^(x-1)+7)/(3^(x-1)+1) = 84</p><p>(9^(x-1)+7)/(3^(x-1)+1) = 4</p><p><strong>Step 4: Solve for x</strong></p><p>Let y = 3^(x-1). Then 9^(x-1) = y²</p><p>(y² + 7)/(y + 1) = 4</p><p>y² + 7 = 4y + 4</p><p>y² - 4y + 3 = 0</p><p>(y - 1)(y - 3) = 0</p><p>So y = 1 or y = 3</p><p><strong>Step 5: Find x values</strong></p><p>If 3^(x-1) = 1, then x - 1 = 0, so <strong>x = 1</strong></p><p>If 3^(x-1) = 3, then x - 1 = 1, so <strong>x = 2</strong></p><p>∴ Answer: x = 1 and x = 2 (AD - both values)</p>
Correct Answer: AD