## Intermediate Algebra (12th Edition)

According to the negative exponent rule, $a^{-n}=\frac{1}{a^{n}}$. Therefore, $\frac{1}{4^{-2}}=\frac{1}{\frac{1}{4^{2}}}=1\div\frac{1}{4^{2}}=1\times4^{2}=1\times16=16$.