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

Let $\mathrm{S}={\log }_{9^{1/2}}\mathrm{x}+{\log }_{9^{1/3}}\mathrm{x}+\dots$

Using property, ${\log }_{{\mathrm{a}}^{\mathrm{b}}}\mathrm{x}=\frac{1}{\mathrm{b}}{\log }_{\mathrm{a}}\mathrm{x}$

$\begin{array}{c}\mathrm{S}=2{\log }_9\mathrm{x}+3{\log }_9\mathrm{x}+\dots +22{\log }_9\mathrm{x}\\ ={\log }_9\mathrm{x}(2+3+4+\dots +22)\\ ={\log }_9\mathrm{x}\left[\frac{21}{2}(4+20)\right]={\log }_9\mathrm{x}(21\times 12)\end{array}$

$\therefore \mathrm{S}=252{\log }_9\mathrm{x}$

Given, S = 504, then

$\begin{array}{l}252 {\log }_9\mathrm{x}=504\\ \Rightarrow {\log }_9\mathrm{x}=2\\ \Rightarrow \mathrm{x}=(9{)}^2=81\end{array}$