1. We are given the equation $$4^p \times 32^{p+4} \div 2^6 = 1$$ and asked to solve for $p$. 2. First, express all terms with the same base. Note that $4 = 2^2$ and $32 = 2^5$. 3. Rewrite the equation: $$4^p = (2^2)^p = 2^{2p}$$ $$32^{p+4} = (2^5)^{p+4} = 2^{5(p+4)} = 2^{5p + 20}$$ 4. Substitute back: $$2^{2p} \times 2^{5p + 20} \div 2^6 = 1$$ 5. Combine the powers of 2 in the numerator: $$2^{2p + 5p + 20} \div 2^6 = 2^{7p + 20} \times 2^{-6} = 2^{7p + 14}$$ 6. So the equation becomes: $$2^{7p + 14} = 1$$ 7. Since $1 = 2^0$, equate exponents: $$7p + 14 = 0$$ 8. Solve for $p$: $$7p = -14$$ $$p = \frac{-14}{7}$$ $$p = -2$$ **Final answer:** $$p = -2$$