${(0.05)^{{{\log }_{_{\sqrt {20} }}}(0.1 + 0.01 + 0.001 + ......)}}= . .$ . .
$81$
${1 \over {81}}$
$20$
$0.05$
જો ${a^2} + 4{b^2} = 12ab $ તો $\log (a + 2b)= . . .$ .
$\sum\limits_{n = 1}^n {{1 \over {{{\log }_{{2^n}}}(a)}}} = $
${81^{(1/{{\log }_5}3)}} + {27^{{{\log }_{_9}}36}} + {3^{4/{{\log }_{_7}}9}} = . . . .$
જો $a = {\log _{24}}12,\,b = {\log _{36}}24$ અને $c = {\log _{48}}36$ તો $1+abc = . . . .$
જો ${\log _{10}}x = y,$ તો ${\log _{1000}}{x^2}= . . .$ .