## Precalculus (6th Edition) Blitzer

$\dfrac{1}{1024}$
RECALL: (i) $a^{\frac{m}{n}}=(\sqrt[n]{a})^m.$ (ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0.$ Use rule (ii) above to obtain $\dfrac{1}{16^{\frac{5}{2}}}.$ Use rule (i) with m=5 and n=2 to obtain $\dfrac{1}{(\sqrt{16})^5} \\=\dfrac{1}{(\sqrt{4^2})^5} \\=\dfrac{1}{(4)^5} \\=\dfrac{1}{1024}.$