$\sum\limits_{n = 1}^n {{1 \over {{{\log }_{{2^n}}}(a)}}} = $
${{n(n + 1)} \over 2}{\log _a}2$
${{n(n + 1)} \over 2}{\log _2}a$
${{{{(n + 1)}^2}{n^2}} \over 4}{\log _2}a$
None of these
Solution set of equation
$\left| {1 - {{\log }_{\frac{1}{6}}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{\frac{1}{6}}}x + {{\log }_{\frac{1}{2}}}x} \right|$ is $\left[ {\frac{a}{b},a} \right],a,b, \in N,$ then the value of $(a + b)$ is
The number of solution $(s)$ of the equation $log_7(2^x -1) + log_7(2^x -7) = 1$, is -
If ${x_n} > {x_{n - 1}} > ... > {x_2} > {x_1} > 1$ then the value of ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}.....{\log _{{x_n}}}{x_n}^{x_{n - 1}^{{ {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} ^{{x_1}}}}}$ is equal to
The number ${\log _{20}}3$ lies in
Logarithm of $32\root 5 \of 4 $ to the base $2\sqrt 2 $ is