If $A = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 \,}}\,2,$ then $A$ is equal to

If ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x,$ then $x$ be

$\sum\limits_{n = 1}^n {{1 \over {{{\log }_{{2^n}}}(a)}}} = $

Let $x, y$ be real numbers such that $x>2 y>0$ and $2 \log (x-2 y)=\log x+\log y$  Then, the possible value(s) of $\frac{x}{y}$

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 ${2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5$ is