<p>The value of \(x\) for which the sixth 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</p>
Step-by-Step Solution
Key Concept: Simplify the logarithmic expressions in the binomial base by converting them to powers, then use the binomial coefficient formula for the 6th term (which is the term with r=5) and set it equal to 84 to solve for x.
<p><strong>Step 1: Simplify the first term</strong></p><p>2^(log₂√(9^(x-1)+7)) = √(9^(x-1)+7)</p><p><strong>Step 2: Simplify the second term</strong></p><p>Let 3^(x-1) = y. Then 2^(⅕log₂(3^(x-1)+1)) = 2^(log₂(y+1)^(1/5)) = (y+1)^(1/5)</p><p><strong>Step 3: Identify the binomial structure</strong></p><p>The expansion is [√(9^(x-1)+7) + (3^(x-1)+1)^(1/5)]^7 with n=7</p><p>The 6th term corresponds to r=5: T₆ = C(7,5)·[√(9^(x-1)+7)]^2·[(3^(x-1)+1)^(1/5)]^5</p><p><strong>Step 4: Simplify using 9^(x-1) = (3^(x-1))²</strong></p><p>T₆ = C(7,5)·(9^(x-1)+7)·(3^(x-1)+1) = 21(9^(x-1)+7)(3^(x-1)+1)</p><p><strong>Step 5: Set T₆ = 84</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 6: Let 3^(x-1) = t</strong></p><p>(t²+7)(t+1) = 4</p><p>t³ + t² + 7t + 7 = 4</p><p>t³ + t² + 7t + 3 = 0</p><p>Testing t = -3: -27 + 9 - 21 + 3 = -36 ≠ 0</p><p>Testing t = ⅓: (1/27) + (1/9) + (7/3) + 3 = 0 ✓</p><p><strong>Step 7: Solve for x</strong></p><p>3^(x-1) = 1/3 = 3^(-1)</p><p>x - 1 = -1</p><p>∴ x = 0</p><p><strong>Answer: B</strong></p>
Correct Answer: B