1. **State the problem:** Solve the equation $2x^2 = \frac{16}{x}$ for $x$. 2. **Rewrite the equation:** Multiply both sides by $x$ to eliminate the denominator: $$x \cdot 2x^2 = x \cdot \frac{16}{x}$$ This gives: $$2x^3 = 16$$ 3. **Simplify the equation:** Divide both sides by 2: $$\frac{\cancel{2}x^3}{\cancel{2}} = \frac{16}{2}$$ $$x^3 = 8$$ 4. **Solve for $x$:** Take the cube root of both sides: $$x = \sqrt[3]{8}$$ 5. **Evaluate the cube root:** $$x = 2$$ 6. **Check for extraneous solutions:** Since we multiplied by $x$, we must ensure $x \neq 0$ (to avoid division by zero in the original equation). $x=2$ is valid. **Final answer:** $$x = 2$$