The real root of the equation ${\log _4}\{ {\log _2}(\sqrt {x + 8} - \sqrt x )\} = 0$ is..........

Let $a, b, x$ be positive real numbers with $a \neq 1, x \neq 1, ab \neq 1$. Suppose $\log_{a} b = 10$,and $\frac{\log_{a} x \cdot \log_{x}(\frac{b}{a})}{\log_{x} b \cdot \log_{ab} x} = \frac{p}{q}$,where $p$ and $q$ are positive integers which are coprime. Then $p+q$ is

If $a, b, c \neq 0$ and belong to the set $\{0, 1, 2, 3, \ldots, 9\}$,then $\log _{10}\left(\frac{a+10 b+10^2 c}{10^{-4} a+10^{-3} b+10^{-2} c}\right)$ is equal to

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

$a^{m \log_a n} = ?$

If $2 \log (x+1)-\log (x^{2}-1)=\log 2$,then $x=$