Question

If the sum of the first $20$ terms of the series ${\log }_{\left(7^{1/2}\right)}x+{\log }_{\left(7^{1/3}\right)}x+{\log }_{\left(7^{1/4}\right)}x+...$ is $460$, then $x$ is equal to :

• A. $7^{1/2}$
• B. $7^2$
• C. $e^2$
• D. $7^{46/21}$

Answer 1

The correct option is B $7^2$

$(2+3+4+...+21){\log }_{7}x=460$
$\Rightarrow (\frac{21\times 22}{2}-1){\log }_{7}x=460$
$\Rightarrow 230{\log }_{7}x=460$
$\Rightarrow {\log }_{7}x=2$
$\Rightarrow \log x=2\log 7$
$\Rightarrow x=7^2$