Answer: 91

Small Hint:

Write $x_n = 3^{a+bn}$ with integers $a$ and $b \ge 1;$ the log-sum condition gives $2a + 7b = 77$

Big Hint:

The sum $x_0 + \cdots + x_7$ lies strictly between $x_7$ and $3 x_7,$ so its base-$3$ log pins down $a + 7b$

Solution:

Every term is a power of $3$ and the ratio is a quotient of powers of $3,$ so $x_n = 3^{a + bn}$ for integers $a$ and $b,$ with $b \ge 1$ since the sequence increases. The first condition gives $$\sum_{n=0}^{7} (a + bn) = 8a + 28b = 308,$$ i.e. $$2a + 7b = 77.$$

For the second condition, $x_7 = 3^{a+7b}$ is the largest term, and $$\begin{aligned} x_7 &\lt \sum_{n=0}^{7} x_n \\ &\lt x_7\left(1 + \tfrac{1}{3} + \tfrac{1}{9} + \cdots\right) \\ &= \tfrac{3}{2}\,x_7 \lt 3x_7, \end{aligned}$$ so $\log_3\left(\sum x_n\right)$ lies strictly between $a + 7b$ and $a + 7b + 1.$ The given bounds then force $a + 7b = 56.$

Subtracting from $2a + 7b = 77$ yields $a = 21,$ then $b = 5.$ Therefore $\log_3(x_{14}) = a + 14b = 21 + 70$ $= 91.$