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

If ${1 \over 2} \le {\log _{0.1}}x \le 2$ then

Let $a=3 \sqrt{2}$ and $b=\frac{1}{5^{\frac{1}{6}} \sqrt{6}}$. If $x, y \in R$ are such that  $3 x+2 y=\log _a(18)^{\frac{5}{4}} \text { and }$  $2 x-y=\log _b(\sqrt{1080}),$  then $4 x+5 y$ is equal to. . . . 

The set of real values of $x$ satisfying ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$ is

The number of real values of the parameter $k$ for which ${({\log _{16}}x)^2} - {\log _{16}}x + {\log _{16}}k = 0$ with real coefficients will have exactly one solution is

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 $................$.