The value of ${(0.05)^{{{\log }_{_{\sqrt {20} }}}(0.1 + 0.01 + 0.001 + ......)}}$ is
$81$
${1 \over {81}}$
$20$
$0.05$
If ${\log _{10}}x = y,$ then ${\log _{1000}}{x^2} $ is equal to
If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)$ then $x \ne 1$ lies in
The set of real values of $x$ for which ${\log _{0.2}}{{x + 2} \over x} \le 1$ is
The number of real values of the parameter $k$ for which ${({\log _{16}}x)^2} - {\log _{16}}x + {\log _{16}}k = 0$ with real coefficients will have exactly one solution is
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