Lovely Logarithm!

Let $a_0 = 3^a$ and $a_n = 3^a3^{nr} = 3^{a+nr}$ , where $a$ and $r$ are integers.

$\begin{aligned} \Rightarrow \displaystyle \sum_{n=0}^7 {\log_3{\left(x_n\right)}} & = \sum_{n=0}^7 {\log_3{\left(3^{a+nr}\right)}} = \sum_{n=0}^7 {\left( a+nr\right)} \\ & = 8a + 28r = 308 \\ \\ \Rightarrow 2a + 7r & = 77 \quad ...(1) \\ \\ \Rightarrow \sum_{n=0}^7 {x_n} & = \sum_{n=0}^7 {3^{a+nr}} = 3^a\dot{} \frac {3^{8r}-1} {3^r-1} \approx 3^{a+7r} \\ \Rightarrow 56 & \le \log_3 {\left(\sum_{n=0}^7 {x_n}\right)} \le 57 \\ 56 & \le \log_3 {\left(3^{a+7r}\right)} \le 57 \\ \\ \Rightarrow 56 & \le a+7r \le 57 \quad ...(2) \end{aligned}$

From Eq.1 $-$ Eq.2:

$20 \le a \le 21 \quad \Rightarrow \begin{cases} a = 20 & \Rightarrow 40+7r = 77 & \Rightarrow r = \frac{37}{7} \text{ rejected}\\ a = 21 & \Rightarrow 42+7r = 77 & \Rightarrow r = 5 \end{cases}$

$\Rightarrow \log_3 {x_{14}} = \log_3 {3^{a+14r}} = a + 14r = 21+14\times 5 = \boxed{91}$

Because of the second condition and given that the series is increasing geometric and consisting of integral powers of 3, it is quite trivial that $x_{7} = 3^{56}$ .

If the factor of the series can be written as $3^{q}$ , with $q \in \mathbb{N}$ , we can see that $x_{6} = 3^{56 -q}$ and $x_{5} = 3^{56 - 2q}$ and so on ...

Then because of the first condition we get the following equation in q :

$56 + 56- q + 56 -2q + 56 -3q + 56 -4q + 56- 5q + 56 -6q + 56 - 7q = 308$

or $8. 56 -28q = 308$

or $q = 5$ .

Therefor the geometric series starts at $x_{0} = 3^{56 - 7q} = 3^{21}$

and $x_{14} = 3^{21} . 3^{14.q} = 3^{21} . 3^{70} = 3^{91}$ , so that the answer is 91.