1. **State the problem:** Solve the equation $$2\left(\frac{1}{8}\right) = 32^{n-1}$$ for $n$. 2. **Rewrite the equation:** Simplify the left side: $$2 \times \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$$ So the equation becomes: $$\frac{1}{4} = 32^{n-1}$$ 3. **Express bases as powers of 2:** Note that: $$\frac{1}{4} = 4^{-1} = (2^2)^{-1} = 2^{-2}$$ and $$32 = 2^5$$ So: $$32^{n-1} = (2^5)^{n-1} = 2^{5(n-1)}$$ 4. **Set the exponents equal:** Since the bases are the same (base 2), the exponents must be equal: $$-2 = 5(n-1)$$ 5. **Solve for $n$:** $$-2 = 5n - 5$$ Add 5 to both sides: $$-2 + 5 = 5n$$ $$3 = 5n$$ Divide both sides by 5: $$n = \frac{3}{5}$$ **Final answer:** $$n = \frac{3}{5}$$