1. **State the problem:** Solve the equation $4^x = \frac{32\sqrt{2}}{2}$ for $x$. 2. **Rewrite the right side:** Simplify the right side: $$\frac{32\sqrt{2}}{2} = 16\sqrt{2}$$ 3. **Express both sides with base 2:** Recall that $4 = 2^2$, so: $$4^x = (2^2)^x = 2^{2x}$$ Also, express $16\sqrt{2}$ as a power of 2: $$16 = 2^4, \quad \sqrt{2} = 2^{\frac{1}{2}}$$ Therefore: $$16\sqrt{2} = 2^4 \times 2^{\frac{1}{2}} = 2^{4 + \frac{1}{2}} = 2^{\frac{9}{2}}$$ 4. **Set the exponents equal:** Since the bases are the same and nonzero, $$2^{2x} = 2^{\frac{9}{2}} \implies 2x = \frac{9}{2}$$ 5. **Solve for $x$:** $$x = \frac{9}{4}$$ **Final answer:** $$\boxed{\frac{9}{4}}$$