1. **State the problem:** Solve the equation $\left(\frac{1}{8}\right)^{2x} = 4 \sqrt{2}$ for $x$. 2. **Rewrite the bases as powers of 2:** Since $8 = 2^3$, we have $\frac{1}{8} = 2^{-3}$. So, $\left(\frac{1}{8}\right)^{2x} = (2^{-3})^{2x} = 2^{-6x}$. 3. **Rewrite the right side as a power of 2:** $4 = 2^2$ and $\sqrt{2} = 2^{\frac{1}{2}}$, so $4 \sqrt{2} = 2^2 \times 2^{\frac{1}{2}} = 2^{2 + \frac{1}{2}} = 2^{\frac{5}{2}}$. 4. **Set the exponents equal:** Since the bases are the same and nonzero, we can equate the exponents: $$-6x = \frac{5}{2}$$ 5. **Solve for $x$:** $$x = -\frac{5}{2} \times \frac{1}{6} = -\frac{5}{12}$$ **Final answer:** $$x = -\frac{5}{12}$$