If $A = \log _2 \log _2 \log _4 256 + 2 \log _{\sqrt{2}} 2$,then $A$ is equal to

The solution set of the inequality ${\log _{10}}({x^2} - 2x - 2) \le 0$ is:

If $\log _k x \cdot \log _5 k = \log _x 5$,where $k \neq 1$ and $k > 0$,then the value of $x$ is:

Given that $a, b \in \{0, 1, 2, \ldots, 9\}$ with $a+b \neq 0$ and that $\left(a+\frac{b}{10}\right)^x = \left(\frac{a}{10}+\frac{b}{100}\right)^y = 1000$. Then,$\frac{1}{x} - \frac{1}{y}$ is equal to

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

$A$ real root of the equation $\log_{4}\{\log_{2}(\sqrt{x + 8} - \sqrt{x})\} = 0$ is