MCQ
The value of ${(0.05)^{{{\log }_{_{\sqrt {20} }}}(0.1 + 0.01 + 0.001 + ......)}}$ is
- ✓$81$
- B${1 \over {81}}$
- C$20$
- D$0.05$
$ = {20^{ - 2{{\log }_{20}}(1/9)}} = {20^{2{{\log }_{20}}9}} = {20^{{{\log }_{20}}{9^2}}} = {9^2} = 81$.
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