A possible value of 'x', for which the ninth term in the expansion of { 3^(log ₃√(25ˣ⁻¹ + 7)) + 3^(( - 1/8 )log ₃(5ˣ⁻¹ + 1)) }¹⁰ in the increasing powers of 3^(( - 1/8 )log ₃(5ˣ⁻¹ + 1)) is equal to 180, is :

Solution

Maths sir · 4 steps

Peel off each 3^(log₃) first, and the ninth term becomes a simple quadratic in 5ˣ⁻¹.

1. Simplify both terms. Use 3^(log₃ y) = y. The first term is a = √(25ˣ⁻¹+7). The second is b = 3^(log₃ (5ˣ⁻¹+1)^(-1/8)) = (5ˣ⁻¹+1)^(-1/8).
2. Pick the ninth term. Increasing powers of b means Tᵣ₊₁ = C(10, r) a¹⁰⁻ʳ bʳ. The ninth term has r = 8, so a is squared and b is raised to the 8th power. That removes both the root and the -1/8.

   T₉ = C(10, 8) a² b⁸ = 45 · (25ˣ⁻¹+7)/(5ˣ⁻¹+1)

3. Set it to 180. Let t = 5ˣ⁻¹, so 25ˣ⁻¹ = t². Then (t²+7)/(t+1) = 4, which gives t² - 4t + 3 = 0, so t = 1 or t = 3.
4. So. t = 1 means 5ˣ⁻¹ = 1, so x = 1. The other root gives x = 1 + log₅ 3, which is not among the options. The answer is (d) 1.

The idea to keep. In binomial questions with powers of 3 and logs inside, first rewrite 3^(log₃ y) as y. The ninth term has r = 8, and a substitution like t = 5ˣ⁻¹ usually turns the condition into a quadratic.

Answer from the NTA final answer key. Solution by PrepWiser.