If $\log_e \left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_e a + \log_e b)$,then the relationship between $a$ and $b$ is:

Suppose $\log _a b + \log _b a = c$. The smallest possible integer value of $c$ for all $a, b > 1$ is

The value of $\left(\left(\log _2 9\right)^2\right)^{\frac{1}{\log _2\left(\log _2 9\right)}} \times(\sqrt{7})^{\frac{1}{\log _4 7}}$ is . . . . . . .

If $\log _3 x+\log _3 y=2+\log _3 2$ and $\log _3(x+y)=2$, then

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 logarithm of $32\sqrt[5]{4}$ to the base $2\sqrt{2}$ is: