The set of real values of $x$ for which ${2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5$ is

If ${\log _{10}}2 = 0.30103,{\log _{10}}3 = 0.47712,$ the number of digits in ${3^{12}} \times {2^8} $ is

If ${x^{{3 \over 4}{{({{\log }_3}x)}^2} + {{\log }_3}x - {5 \over 4}}} = \sqrt 3 $ then $x$ has

Solution set of equation

$\left| {1 - {{\log }_{\frac{1}{6}}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{\frac{1}{6}}}x + {{\log }_{\frac{1}{2}}}x} \right|$ is $\left[ {\frac{a}{b},a} \right],a,b, \in N,$ then the value of $(a + b)$ is

Let $\left(x_0, y_0\right)$ be the solution of the following equations $(2 x)^{\ln 2} =(3 y)^{\ln 3}$ $3^{\ln x} =2^{\ln y}$ . Then $x_0$ 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 $................$.