Let $\quad \sum \limits_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c$, where $a, b, c \in Z$ and $e=\sum \limits_{n=0}^{\infty} \frac{1}{n!}$ Then $a^2-b+c$ is equal to $................$.

If ${\log _4}5 = a$ and ${\log _5}6 = b,$ then ${\log _3}2$ is equal to

The number of solution of ${\log _2}(x + 5) = 6 - x$ is

For $y = {\log _a}x$ to be defined $'a'$ must be

If $log_ab + log_bc + log_ca$ vanishes where $a, b$ and $c$ are positive reals different than unity then the value of $(log_ab)^3 + (log_bc)^3 + (log_ca)^3$ is

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