Lovely Logarithm!

n=07log3(xn)=30856log3(n=07xn)57 \begin{aligned} &\displaystyle\sum_{n=0}^{7}\log_{3}(x_{n}) &= 308 \\ 56 \leq & \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right )& \leq 57 \\ \end{aligned}

The increasing geometric sequence x0,x1,x2,x_{0},x_{1},x_{2},\ldots consists entirely of integral powers of 3. If they satisfy the two conditions above, find log3(x14).\log_{3}(x_{14}).

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