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

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(D) The series $x = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ is an infinite geometric progression with first term $a_1 = \frac{1}{4}$ and com

Vedclass · Accepted Answer

$4$

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(D) The series $x = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ is an infinite geometric progression with first term $a_1 = \frac{1}{4}$ and common ratio $r = \frac{1}{2}$.Using the sum formula $S_{\infty} = \frac{a_1}{1 - r}$,we get $x = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.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 have $\log_{5^{1/2}} (2^{-1}) = \frac{1}{1/2} \log_5 (2^{-1}) = 2 \log_5 (2^{-1}) = \log_5 (2^{-2}) = \log_5 (1/4)$.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:

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