Let
S=log91/2x+log91/3x+log91/4x+…..+log91/22x
⇒S=2log9x+3log9x+……+22log9x
⇒S=log9x(2+3+…..+22)
⇒S=log9x221(2+22)
⇒S=252log9x
Given,
S=504
⇒252log9x=504
⇒log9x=2
⇒x=81
JEE Main 2021 — Mathematics Algebra
If sum of the first 21 terms of the series log91/2x+log91/3x+log91/4x+….. where x>0 is 504, then x is equal to
Held on 20 Jul 2021 · Verified 6 Jul 2026.
243
9
7
81
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