If ${\log _{12}}27 = a,$ then ${\log _6}16 = $

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

Which is the correct order for a given number $\alpha $in increasing order

If $a, b, c$ are distinct positive numbers, each different from $1$, such that $[{\log _b}a{\log _c}a - {\log _a}a] + [{\log _a}b{\log _c}b - {\log _b}b]$ $ + [{\log _a}c{\log _b}c - {\log _c}c] = 0,$ then $abc =$

${\log _4}18$ is

The number of integral solutions $x$ of $\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^2 \geq 0$ is