{(0.05)^{{{log }_{_{sqrt {20} }}}(0.1 + 0.01 | vedclass

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Question

(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}}} \righ

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(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$.

${(0.05)^{{{\log }_{_{\sqrt {20} }}}(0.1 + 0.01 + 0.001 + ......)}}= . .$ . .

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