$7\log \left( {{{16} \over {15}}} \right) + 5\log \left( {{{25} \over {24}}} \right) + 3\log \left( {{{81} \over {80}}} \right)$ is equal to

If $x = {\log _a}(bc),y = {\log _b}(ca),z = {\log _c}(ab),$then which of the following is equal to $1$

If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1),$ then $x$ lies in the interval

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

The sum $\sum \limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}$ is equal to :

The value of $6+\log _{\frac{3}{2}}\left(\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \ldots}}}\right)$ is