## Precalculus (6th Edition)

$\color{blue}{\dfrac{1}{y^{10}}}$
RECALL: (1) $a^{-m} = \dfrac{1}{a^m}$ (2) $\dfrac{a^m}{a^n} = a^{m-n}$ (3) $\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$ (4) $(ab)^m = a^mb^m$ (5) $(a^m)^n=a^{mn}$ (6) $a^m \cdot a^n = a^{m+n}$ (7) $a^0=1, a\ne0$ Use rule (4) above to obtain: $=\dfrac{(8^{-4}(y^2)^{-4}) \cdot (8^{-2}(y^5)^{-2})}{(8^{-3})^{2}(y^{-4})^2}$ Use rule (5) above to obtain: $=\dfrac{8^{-4}y^{2(-4)}\cdot (8^{-2}(y^{5(-2)})}{8^{-3(2)}y^{-4(2)}} \\=\dfrac{8^{-4}y^{-8} \cdot 8^{-2}y^{-10}}{8^{-6}y^{-8}}$ Use rule (6) above to obtain: $=\dfrac{8^{-4+(-2)}y^{-8+(-10)}}{8^{-6}y^{-8}} \\=\dfrac{8^{-6}y^{-18}}{8^{-6}y^{-8}}$ Use rule (2) above to obtain: $=8^{-6-(-6)}y^{-18-(-8)} \\=8^{-6+6}y^{-18+8} \\=8^0y^{-10}$ Use rule (7) to obtain: $\\=1\cdot y^{-10} \\=y^{-10}$ Use rule (1) above to obtain: $\\=\color{blue}{\dfrac{1}{y^{10}}}$