$\sum\limits_{r = 1}^{89} {{{\log }_3}(\tan \,\,{r^o})} = $
$3$
$1$
$2$
$0$
If $x, y, z \in R^+$ are such that $z > y > x > 1$ , ${\log _y}x + {\log _x}y = \frac{5}{2}$ and ${\log _z}y + {\log _y}z = \frac{{10}}{3}$ then ${\log _x}z$ is equal to
If ${a^2} + 4{b^2} = 12ab,$ then $\log (a + 2b)$ is
The number ${\log _{20}}3$ lies in
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
If ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x,$ then $x$ be