1. **State the problem:** Solve the exponential equation $$2^{3x-1} = \frac{\sqrt{2}}{8}$$. 2. **Recall the properties of exponents and radicals:** - $\sqrt{2} = 2^{\frac{1}{2}}$ - $8 = 2^3$ 3. **Rewrite the right side with base 2:** $$\frac{\sqrt{2}}{8} = \frac{2^{\frac{1}{2}}}{2^3} = 2^{\frac{1}{2} - 3} = 2^{-\frac{5}{2}}$$ 4. **Set the exponents equal since the bases are the same:** $$3x - 1 = -\frac{5}{2}$$ 5. **Solve for $x$:** $$3x = -\frac{5}{2} + 1 = -\frac{5}{2} + \frac{2}{2} = -\frac{3}{2}$$ 6. **Divide both sides by 3:** $$x = \frac{-\frac{3}{2}}{3} = -\frac{3}{2} \times \frac{1}{3} = -\frac{3}{2} \times \frac{1}{3} = -\frac{1}{2}$$ **Final answer:** $$x = -\frac{1}{2}$$