smc

This problem is quickly solved with knowledge of the laws of exponents and logarithms. $$\begin{array}{rl}{7}^{x+7}& =8^x\\ 7^x\ast 7^7& =8^x\\ {\left(\frac{8}{7}\right)}^x& =7^7\\ x& ={\log }_{8/7}{7}^7\end{array}$$ Since we are looking for the base of the logarithm, our answer is $\boxed{\frac{8}{7}}$.