The value of $\log _{3}4 \cdot \log _{4}5 \cdot \log _{5}6 \cdot \log _{6}7 \cdot \log _{7}8 \cdot \log _{8}9$ is

The solution set of the equation $x^{\log_x(1-x)^2} = 9$ is:

The sum of the roots of the equation $x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$ 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 ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$,then in which interval does $x$ lie?

If $\frac{1}{\log_3 \pi} + \frac{1}{\log_4 \pi} > x$,then $x$ can be: