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

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 $x$.

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

If $n = (1999)!$,then $\sum\limits_{x = 1}^{1999} {{\log }_n x}$ is equal to

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

Consider all natural numbers whose decimal expansion has only the even digits $0, 2, 4, 6, 8$. Suppose these are arranged in increasing order. If $a_n$ denotes the $n$-th number in this sequence,then $\frac{\lim_{n \rightarrow \infty} \log a_n}{\log n}$ equals