If the sum of the first 20 terms of the serie | vedclass

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(D) The given series is $\sum_{n=2}^{21} \log_{7^{1/n}} x = 460$. Using the property $\log_{a^k} b = \frac{1}{k} \log_a b$,we get $\log_{7^{1/n}} x =

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(D) The given series is $\sum_{n=2}^{21} \log_{7^{1/n}} x = 460$. Using the property $\log_{a^k} b = \frac{1}{k} \log_a b$,we get $\log_{7^{1/n}} x = n \log_7 x$. Thus,the sum is $\sum_{n=2}^{21} n \log_7 x = 460$. $\log_7 x \cdot (2 + 3 + 4 + \dots + 21) = 460$. The sum of the arithmetic progression $2 + 3 + \dots + 21$ is $\frac{20}{2} (2 + 21) = 10 	imes 23 = 230$. So,$230 \cdot \log_7 x = 460$. $\log_7 x = 2$. Therefore,$x = 7^2 = 49$.

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

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