## Calculus with Applications (10th Edition)

$\dfrac{1}{64}$
RECALL: (i) $\dfrac{a^m}{a^n} = a^{m-n}$ (ii) $a^{-m} = \dfrac{1}{a^m}, a \ne 0$ Use rule (i) above to have: $=4^{-2-1} \\=4^{-3}$ Use rule (ii) above to have: $=\dfrac{1}{4^3} \\=\dfrac{1}{64}$