1. The problem asks to express the logarithmic equation $\log(\frac{1}{8}) = -1$ in its equivalent exponential form. 2. Recall that $\log_b(a) = c$ is equivalent to $b^c = a$. 3. In this problem, the base is assumed to be 10 (common logarithm) unless otherwise specified. 4. Using the equivalence, rewrite $\log(\frac{1}{8}) = -1$ as: $$10^{-1} = \frac{1}{8}$$ 5. This means that $10^{-1}$ equals the number $\frac{1}{8}$. 6. Notice that $10^{-1} = \frac{1}{10}$, which is not equal to $\frac{1}{8}$, so the equation as stated is not true if the base is 10. 7. If the logarithm base is changed to 8, then $\log_8(\frac{1}{8}) = -1$ since: $$8^{-1} = \frac{1}{8}$$ 8. Therefore, the logarithmic equation $\log_8(\frac{1}{8}) = -1$ in exponential form is $8^{-1} = \frac{1}{8}$. Final answer: $$8^{-1} = \frac{1}{8}$$