If the sum of the first 21 terms of the series _9^1/2x+_9^1/3x+_9^1/4x+, where x0, is 504…

Question

If the sum of the first $21$ terms of the series $\log_{9^{1/2}}x+\log_{9^{1/3}}x+\log_{9^{1/4}}x+\cdots$, where $x>0$, is $504$, then $x$ is equal to:

Accepted Answer

$$s = 2{\log _9}x + 3{\log _9}x + ....... + 22{\log _9}x$$

$$s = {\log _9}x(2 + 3 + ..... + 22)$$

$$s = {\log _9}x\left\{ {{{21} \over 2}(2 + 22)} \right\}$$

Given, $$252{\log _9}x = 504$$

$$ \Rightarrow {\log _9}x = 2 \Rightarrow x = 81$$

Solutions · 1