The value of (0.05)^{log_{sqrt{20}}(0.1 + 0.0 | vedclass

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

Question

(A) Let $S = 0.1 + 0.01 + 0.001 + \dots$ This is an infinite geometric series with first term $a = 0.1$ and common ratio $r = 0.1$. The sum $S = \frac

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(A) Let $S = 0.1 + 0.01 + 0.001 + \dots$ This is an infinite geometric series with first term $a = 0.1$ and common ratio $r = 0.1$. The sum $S = \frac{a}{1-r} = \frac{0.1}{1-0.1} = \frac{0.1}{0.9} = \frac{1}{9}$. Now,the expression is $(0.05)^{\log_{\sqrt{20}}(1/9)}$. Note that $0.05 = \frac{5}{100} = \frac{1}{20}$. Also,$\sqrt{20} = 20^{1/2}$. Using the property $\log_{a^n}(b) = \frac{1}{n} \log_a(b)$,we have $\log_{20^{1/2}}(1/9) = \frac{1}{1/2} \log_{20}(1/9) = 2 \log_{20}(1/9) = \log_{20}(1/9)^2 = \log_{20}(1/81)$. So,the expression becomes $(1/20)^{\log_{20}(1/81)}$. Using the property $a^{\log_a(x)} = x$,we rewrite $(1/20)^{\log_{20}(1/81)}$ as $(20^{-1})^{\log_{20}(1/81)} = (20^{\log_{20}(1/81)})^{-1} = (1/81)^{-1} = 81$.

The value of $(0.05)^{\log_{\sqrt{20}}(0.1 + 0.01 + 0.001 + \dots)}$ is

Similar Questions

The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots$ up to the $11^{th}$ term is:

The sum of an infinite geometric series is $3$. $A$ series,which is formed by squaring its terms,also has a sum of $3$. The first series is

If $\alpha_r$ and $\beta_r$ (where $\alpha_r < \beta_r$) are the roots of the quadratic equation $x^2 - r^2(r + 1)x + r^5 = 0$,then find the value of $\sum_{r=1}^{n} (3\alpha_r + 2\beta_r)$.

What is the sum of the series $1^3 - 2^3 + 3^3 - 4^3 + 5^3 - 6^3 + 7^3 - 8^3 + 9^3$?

If $\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots + \infty = \frac{\pi^4}{90}$,then the value of $\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \dots + \infty$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module