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STD 11 - basic of algoritham — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - basic of algoritham4 Marks
MCQ
The value of ${(0.05)^{{{\log }_{_{\sqrt {20} }}}(0.1 + 0.01 + 0.001 + ......)}}$ is
  • ✓
    $81$
  • B
    ${1 \over {81}}$
  • C
    $20$
  • D
    $0.05$
✓

Answer

Correct option: A.
$81$
a
(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$.

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