If $y = \log_x x$, then the value of $\left( \frac{dy}{dx} \right)^2$ at $x = e^2$ 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 real solution is:

If ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$,then in which interval does $x$ lie?

The number $\log_{20} 3$ lies in

$\text{If } \log _{0.2}(x-1) > \log _{0.04}(x+5), \text{ then }$

If $\log _{27}(\log _3 x) = \frac{1}{3}$,then the value of $x$ is