smc

First, note that $\frac{1}{8}=8^{-1}$. That means the expression can be rewritten as $$\log 8÷\log 8^{-1}$$ $$\log 8\cdot \frac{1}{\log 8^{-1}}$$ $$\log 8\cdot \frac{1}{-1\cdot \log 8}$$ This simplifies to $-1$, which is answer choice $\boxed{-1}$.