The solution of the equation $7^{x+7} = 8^x$ can be expressed in the form $x = \log_b 7^7$. What is $b$?
First, let's write the equation $7^{x+7} = 8^x$ in the form $7^x \cdot 7^7 = 8^x$. Then we can divide both sides by $7^x$ to get $7^7 = \left(\frac{8}{7}\right)^x$ Then we are looking for an $x$ that can make $\left(\frac{8}{7}\right)^x$ equal to $7^7$. That means $\left(\frac{8}{7}\right)^x = \left(\frac{8}{7}\right)^{\log_{\frac{8}{7}} 7^7}$, so $x = \log_{\frac{8}{7}} 7^7$. And $\log_b 7^7 = \frac{7 \log_b 7}{\log_b \frac{8}{7}}$, so $b$ is equal to $\frac{8}{7}$.

