The value of {a^{{{log }_b}x}}, where a = 0.2 | vedclass

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Question

(d) $x = \frac{{1/4}}{{1 - (1/2)}} = \frac{1}{2}$

$	herefore $${\left( {\frac{1}{5}} ight)^{{{\log }_{\sqrt 5 }}\left( {\frac{1}{2}} ight)}} =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(d) $x = \frac{{1/4}}{{1 - (1/2)}} = \frac{1}{2}$

$	herefore $${\left( {\frac{1}{5}} ight)^{{{\log }_{\sqrt 5 }}\left( {\frac{1}{2}} ight)}} = {\left( {\frac{1}{5}} ight)^{{{\log }_5}\left( {\frac{1}{4}} ight)}} = {5^{ - {{\log }_5}{4^{ - 1}}}} = {5^{{{\log }_5}4}} = 4$.

The value of ${a^{{{\log }_b}x}}$, where $a = 0.2,\;b = \sqrt 5 ,\;x = \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + .........$to $\infty $ is

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