The value of a^{log_b x},where a = 0.2,b = sq | vedclass

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

Question

(D) First,calculate the sum of the infinite geometric series $x = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$This is a geometric progression wit

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) First,calculate the sum of the infinite geometric series $x = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$This is a geometric progression with first term $a_1 = \frac{1}{4}$ and common ratio $r = \frac{1}{2}$.The sum $x = \frac{a_1}{1 - r} = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.Now,substitute $a = 0.2 = \frac{1}{5}$,$b = \sqrt{5} = 5^{1/2}$,and $x = \frac{1}{2}$ into the expression $a^{\log_b x}$:$a^{\log_b x} = (\frac{1}{5})^{\log_{\sqrt{5}} (1/2)} = (5^{-1})^{\log_{5^{1/2}} (2^{-1})}$.Using the property $\log_{b^n} x = \frac{1}{n} \log_b x$,we get $\log_{5^{1/2}} (2^{-1}) = \frac{1}{1/2} \log_5 (2^{-1}) = 2 \log_5 (2^{-1}) = \log_5 (2^{-1})^2 = \log_5 (2^{-2}) = \log_5 (1/4)$.Thus,the expression becomes $(5^{-1})^{\log_5 (1/4)} = 5^{-\log_5 (1/4)} = 5^{\log_5 (1/4)^{-1}} = 5^{\log_5 4} = 4$.

The value of $a^{\log_b x}$,where $a = 0.2$,$b = \sqrt{5}$,and $x = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ to $\infty$ is:

Similar Questions

If $\omega$ is a non-real root of the equation $x^3 - 1 = 0$,then the value of $\sum_{r=1}^5 (1 + \omega^r + \omega^{2r})$ is

If the harmonic mean between $a$ and $b$ is $H$,then $\frac{H + a}{H - a} + \frac{H + b}{H - b} = $

If the sum of infinite terms of a $G.P.$ is $3$ and the sum of the squares of its terms is $3$,then its first term and common ratio are

In a $G.P.$,the first term is $7$,the last term is $448$,and the sum is $889$. Find the common ratio.

If $\frac{3 + 5 + 7 + \dots \text{ to } n \text{ terms}}{5 + 8 + 11 + \dots \text{ to } 10 \text{ terms}} = 7$,then the value of $n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module