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
$a = b$
$a = {b \over 2}$
$2a = b$
$a = {b \over 3}$
The sum of all the natural numbers for which $log_{(4-x)}(x^2 -14x + 45)$ is defined is -
If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1),$ then $x$ lies in the interval
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
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
The sum $\sum \limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}$ is equal to :