Question

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

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

Answer 1

Answer: (D)

$460=\log _{7} x \cdot(2+3+4+\ldots . .+20+21)$
$\Rightarrow 460=\log _{7} x \cdot\left(\frac{21 \times 22}{2}-1\right)$
$\Rightarrow 460=230 \cdot \log _{7} x$
$\Rightarrow \log _{7} x=2$
$ \Rightarrow x=49$