If ${\log _e}\left( {{{a + b} \over 2}} \right) = {1 \over 2}({\log _e}a + {\log _e}b)$, then relation between $a$ and $b$ will be
$a = b$
$a = {b \over 2}$
$a =2 b$
$a = {b \over 3}$
If ${\log _{12}}27 = a,$ then ${\log _6}16 = $
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 $................$.
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. . . .
If $A = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 \,}}\,2,$ then $A$ is equal to