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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

If $\log _{2}\left(9^{x-1}+7\right)-\log _{2}\left(3^{x-1}+1\right)=2$,then $x$ values are

The equation $x^{(3/4)(\log_2 x)^2 + (\log_2 x) - 5/4} = \sqrt{2}$ has

If $\log 2=a, \log 3=b, \log 7=c$ and $6^x=7^{x+4}$, then $x$ is equal to:

If $x, y, z \in R^+$ are such that $z > y > x > 1$,$\log_{y}x + \log_{x}y = \frac{5}{2}$ and $\log_{z}y + \log_{y}z = \frac{10}{3}$,then $\log_{x}z$ is equal to

If $\frac{1}{\log_3 \pi} + \frac{1}{\log_4 \pi} > x$,then $x =$ ?

Vedclass Test Series

Exam Paper Generator

Online Exam Module