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

The roots of the equation $2^{x + 2} \cdot 27^{x/(x - 1)} = 9$ are given by

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

The solution set of the equation $\left| {1 - {{\log }_{1/6}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{1/6}}x + {{\log }_{1/2}}x} \right|$ is $\left[ {\frac{a}{b},a} \right]$,where $a, b \in N$. Then the value of $(a + b)$ is:

If $y = 2^{1/\log_x 4}$,then $x$ is equal to

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 $x$.