If $a = {\log _{24}}12,\,b = {\log _{36}}24$ and $c = {\log _{48}}36,$ then $1+abc$ is equal to
$2ab$
$2ac$
$2bc$
$0$
The value of $(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . to \infty\right)}$ is equal to
The number of solution $(s)$ of the equation $log_7(2^x -1) + log_7(2^x -7) = 1$, is -
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
If ${a^2} + 4{b^2} = 12ab,$ then $\log (a + 2b)$ is