Question

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

• A. $81$
• B. $243$
• C. $9$
• D. $7$

Answer 1

The correct option is A $81$

${\log }_{9^{\frac{1}{2}}}x+{\log }_{9^{\frac{1}{3}}}x+{\log }_{9^{\frac{1}{4}}}x+...$
$\Rightarrow {\log }_9{x}^2+{\log }_9{x}^3+{\log }_9{x}^4+...$
$\Rightarrow {\log }_9\left(x^{2+3+\cdots 21\text{terms}}\right)=504$
$\Rightarrow 252{\log }_{9}x=504$
$\Rightarrow x=9^2=81$