The increasing [geometric sequence](https://artofproblemsolving.com/wiki/index.php/Geometric_sequence) $x_{0},x_{1},x_{2},\ldots$ consists entirely of [integral](https://artofproblemsolving.com/wiki/index.php/Integer) powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$
Suppose that $x_0 = a$, and that the common [ratio](https://artofproblemsolving.com/wiki/index.php/Ratio) between the terms is $r$. 

The first conditions tells us that $\log_3 a + \log_3 ar + \ldots + \log_3 ar^7 = 308$. Using the rules of [logarithms](https://artofproblemsolving.com/wiki/index.php/Logarithm), we can simplify that to $\log_3 a^8r^{1 + 2 + \ldots + 7} = 308$. Thus, $a^8r^{28} = 3^{308}$. Since all of the terms of the geometric sequence are integral powers of $3$, we know that both $a$ and $r$ must be powers of 3. Denote $3^x = a$ and $3^y = r$. We find that $8x + 28y = 308$. The possible positive integral pairs of $(x,y)$ are $(35,1),\ (28,3),\ (21,5),\ (14,7),\ (7,9),\ (0,11)$.

The second condition tells us that $56 \le \log_3 (a + ar + \ldots ar^7) \le 57$. Using the sum formula for a [geometric series](https://artofproblemsolving.com/wiki/index.php/Geometric_series) and substituting $x$ and $y$, this simplifies to $3^{56} \le 3^x \frac{3^{8y} - 1}{3^y-1} \le 3^{57}$. The fractional part $\approx \frac{3^{8y}}{3^y} = 3^{7y}$. Thus, we need $\approx 56 \le x + 7y \le 57$. Checking the pairs above, only $(21,5)$ is close.

Our solution is therefore $\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = \boxed{091}$.