Evaluate the sum: $\sum\limits_{k = 1}^n \frac{1}{\log_{2^k}(a)}$

The sum of all the natural numbers for which $\log_{(4-x)}(x^2 - 14x + 45)$ is defined is -

If $(x^2 \log _x 27) \cdot \log _9 x = x + 4$, then the value of $x$ is

If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)$,then $x$ lies in which 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 $x$.

If $a^2 + 4b^2 = 12ab$,then $\log(a + 2b) =$