1. The problem asks to express the equation $\frac{\log \frac{1}{8}}{\log 8} = -1$ as an equivalent exponent. 2. Recall that $\log a = b$ means $10^b = a$ if the log is base 10 (common logarithm). Here, the expression involves a ratio of logarithms. 3. Let us set $x = \frac{\log \frac{1}{8}}{\log 8}$ and see what it means. Writing logs explicitly: $$x = \frac{\log \left(8^{-1}\right)}{\log 8}$$ because $\frac{1}{8} = 8^{-1}$. 4. Using logarithmic identity: $\log a^b = b \log a$, so: $$x = \frac{-1 \log 8}{\log 8}$$ 5. Simplify the fraction: $$x = \frac{-1 \log 8}{\log 8} = -1$$ 6. Therefore, the value of the original expression is $-1$. 7. To write this in an exponent form, observe that: $$\frac{\log \frac{1}{8}}{\log 8} = -1$$ means $$\log_{8} \frac{1}{8} = -1$$ 8. By the definition of logarithm base change, this means: $$8^{-1} = \frac{1}{8}$$ which confirms the equation. Final equivalent exponent form: $$\boxed{8^{-1} = \frac{1}{8}}$$