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Basic of Logarithms
hard
${(0.05)^{{{\log }_{_{\sqrt {20} }}}(0.1 + 0.01 + 0.001 + ......)}}= . .$ . .
A
$81$
B
${1 \over {81}}$
C
$20$
D
$0.05$
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