## Precalculus (6th Edition)

RECALL: (1) $a^{-m} = \dfrac{1}{a^m}$ (2) $\left(\dfrac{a}{b}\right)^m = \dfrac{a^m}{b^m}$ Use the rule (1) above to obtain: $(\frac{2}{3})^{-2} = \dfrac{1}{(\frac{2}{3})^2}$ Use rule (2) above to obtain: $=\dfrac{1}{(\frac{2^2}{3^2})}$ Use the rule $a \div \frac{b}{c}=a \cdot \frac{c}{b}$ to obtain: $=1 \cdot \left(\frac{3^2}{2^2}\right) \\=\dfrac{3^2}{2^2}$ Use rule (2) above to obtain: $=\left(\dfrac{3}{2}\right)^2$ Thus, the given statement is true.