If $\log _{2}\left(9^{x-1}+7\right)-\log _{2}\left(3^{x-1}+1\right)=2$,then $x$ values are

$\log (\cosh 3 + \sinh 3) + \log (\cosh 3 - \sinh 3) = $

If ${a^{x - 1}} = bc$,${b^{y - 1}} = ca$,and ${c^{z - 1}} = ab$,then find the value of $\sum \frac{1}{x}$.

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 ${\log _4}5 = a$ and ${\log _5}6 = b,$ then find the value of ${\log _3}2$.

If $\log _k x \cdot \log _5 k = \log _x 5$,where $k \neq 1$ and $k > 0$,then the value of $x$ is: