Value of {(0.05)^{{{log }_{_{√ {20} }}}(0.1 + 0.01 + 0.001 + ......)}} is

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Basic of Logarithms

hard

The value of ${(0.05)^{{{\log }_{_{\sqrt {20} }}}(0.1 + 0.01 + 0.001 + ......)}}$ is

$81$

${1 \over {81}}$

$20$

$0.05$

Solution

(a) ${(0.05)^{{{\log }_{\sqrt {20} }}(0.1 + 0.01 + ……)}} = {\left( {{1 \over {20}}} \right)^{2{{\log }_{20}}\left( {{{0.1} \over {1 – 0.1}}} \right)}}$

$ = {20^{ – 2{{\log }_{20}}(1/9)}} = {20^{2{{\log }_{20}}9}} = {20^{{{\log }_{20}}{9^2}}} = {9^2} = 81$.

Standard 11

Mathematics

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hard

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medium

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normal

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normal

(JEE MAIN-2023)

If ${{\log x} \over {b – c}} = {{\log y} \over {c – a}} = {{\log z} \over {a – b}},$ then which of the following is true