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 $x$.

• A
  $x = a$
• B
  $x = a^a$
• C
  $x = a^{-1/a}$
• D
  $x = a^{1/a}$