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

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 product of all positive real values of $x$ satisfying the equation $x^{\left(16\left(\log _5 x\right)^3-68 \log _5 x\right)}=5^{-16}$is. . . . .

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

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 ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$ then $x$ belongs to the interval