If sum of the first 21 terms of the series log _(9^(1/2))x + log _(9^(1/3))x + log _(9^(1/4))x + ......., where x > 0 is 504, then x is equal to

Solution

Maths sir · 4 steps

Pull the root out of the base: log_(9^(1/n)) x = n log₉ x, and the series becomes a plain AP.

1. Simplify each term. Use log_(aᵏ) x = 1/k logₐ x. With k = 1/n, the term log_(9^(1/n)) x becomes n log₉ x.
2. Spot the AP. The series is 2 log₉ x + 3 log₉ x + 4 log₉ x + …. The 21 terms run from n = 2 to n = 22, so log₉ x is multiplied by 2 + 3 + … + 22.

   2 + 3 + … + 22 = (22 × 23)/2 - 1 = 253 - 1 = 252

3. Solve for x. 252 log₉ x = 504 gives log₉ x = 2, so x = 9² = 81.
4. So. x = 81, which is option (d).

The idea to keep. When the base of a log carries a power, bring it out as a reciprocal: log_(aᵏ) x = 1/k logₐ x. Then the common log factors out and you are left adding plain numbers.

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