If $\log _{0.04}(x - 1) \ge \log _{0.2}(x - 1)$,then $x$ belongs to the interval:

The equation $x^{\frac{3}{4}(\log_2 x)^2 + \log_2 x - \frac{5}{4}} = \sqrt{2}$ has

If $x = \log_{0.1} 0.001$ and $y = \log_9 81$, then $\sqrt{x - 2\sqrt{y}}$ is equal to

The number of solutions of the equation $\log _{4}(x-1)=\log _{2}(x-3)$ is

The value of $x$ satisfying $\log _a x + \log _{\sqrt{a}} x + \log _{\sqrt[3]{a}} x + \dots + \log _{\sqrt[n]{a}} x = \frac{n(n+1)}{2}$ is given by the sum of $n$ terms. If the series is $\log _a x + \log _{a^{1/2}} x + \log _{a^{1/3}} x + \dots + \log _{a^{1/n}} x = S$,find $x$ for the given expression $\log _a x + \log _{\sqrt{a}} x + \dots + \log _{a^{1/n}} x = \frac{n(n+1)}{2}$. For the specific case provided: $\log _a x + \log _{a^{1/2}} x + \dots + \log _{a^{1/n}} x = \sum_{k=1}^{n} k \log_a x = \frac{n(n+1)}{2} \log_a x$. Given $\frac{n(n+1)}{2} \log_a x = \frac{a+1}{2}$,find $x$.

$x = \log \left( \frac{1}{y} + \sqrt{1 + \frac{1}{y^2}} \right) \Rightarrow y$ is equal to