1. **State the problem:** Solve the equation $$16 \cdot 64^{3 - 2r} = \left(\frac{1}{4}\right)^{3r}$$ for $r$. 2. **Rewrite bases as powers of 2:** - $16 = 2^4$ - $64 = 2^6$ - $\frac{1}{4} = 4^{-1} = (2^2)^{-1} = 2^{-2}$ 3. **Rewrite the equation using these powers:** $$2^4 \cdot (2^6)^{3 - 2r} = (2^{-2})^{3r}$$ 4. **Simplify exponents:** $$2^4 \cdot 2^{6(3 - 2r)} = 2^{-2 \cdot 3r}$$ 5. **Combine the left side powers:** $$2^{4 + 18 - 12r} = 2^{-6r}$$ 6. **Simplify the exponent on the left:** $$2^{22 - 12r} = 2^{-6r}$$ 7. **Since bases are equal, set exponents equal:** $$22 - 12r = -6r$$ 8. **Solve for $r$:** $$22 = -6r + 12r$$ $$22 = 6r$$ $$r = \frac{22}{6}$$ 9. **Simplify the fraction:** $$r = \frac{\cancel{22}^{11}}{\cancel{6}^3} = \frac{11}{3}$$ **Final answer:** $$r = \frac{11}{3}$$