If $n = (1999)!$,then $\sum\limits_{x = 1}^{1999} {{\log }_n x}$ is equal to

Let $x$ and $y$ be real numbers such that $x > 2y > 0$ and $2 \log (x - 2y) = \log x + \log y$. Then,the possible value$(s)$ of $\frac{x}{y}$ is/are:

If $a^x = b^y = c^z = d^w$,then the value of $x\left(\frac{1}{y} + \frac{1}{z} + \frac{1}{w}\right)$ is

If $a < 1$ and $2 \operatorname{Sinh}^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\log \left(\frac{1+x}{1-x}\right)$,then $x=$

The solution of the equation $\log _{101} \log _{7}(\sqrt{x+7}+\sqrt{x})=0$ is

What is the number of real values of the parameter $k$ for which the equation $({\log _{16}}x)^2 - {\log _{16}}x + {\log _{16}}k = 0$ has exactly one solution,given that the coefficients are real?