Question

If ${\log }_{\sqrt{7}}x+{\log }_{\sqrt[3]{37}}x+{\log }_{\sqrt[4]{47}}x+\cdots$ upto ${20}^{th}$ term $=230$,then the value of $x$ equals

• A. $7^2$
• B. $7$
• C. $7^3$
• D. None of these

Answer 1

Answer: (B)

$7$

${\log }_{\sqrt{7}}x+{\log }_{\sqrt[3]{37}}x+{\log }_{\sqrt[4]{47}}x+\cdots +{\log }_{\sqrt[21]{217}}x=230$

$\Rightarrow 2{\log }_{7}x+3{\log }_{7}x+4{\log }_{7}x+...+21{\log }_{7}x=230$

$\Rightarrow {\log }_{7}x(2+3+4+....+21)=230$

$\Rightarrow {\log }_{7}x[{\sum }_{n=1}^{21}n-1]=230$

$\Rightarrow {\log }_{7}x[\frac{21\left(22\right)}{2}-1]=230$

$\Rightarrow {\log }_{7}x\left(230\right)=230$

$\Rightarrow x=7$