The logarithm of $32\sqrt[5]{4}$ to the base $2\sqrt{2}$ is:

If $\log _{2} 6 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8)$, then the values of $x$ are

If $a, b, c$ are three distinct positive numbers different from $1$ such that $[(\log_b a)(\log_c a) - \log_a a] + [(\log_a b)(\log_c b) - \log_b b] + [(\log_a c)(\log_b c) - \log_c c] = 0$,then $abc =$

If $x_n > x_{n-1} > \dots > x_2 > x_1 > 1$,then the value of $\log_{x_1} \log_{x_2} \log_{x_3} \dots \log_{x_n} (x_n^{x_{n-1}^{\dots^{x_1}}})$ is:

If $\log x : \log y : \log z = (y - z) : (z - x) : (x - y)$,then

The value of $\log_2(\log_3(\dots(\log_{100}(100^{99^{98^{\dots^{2^1}}})))\dots))}$ is