## Thinking Mathematically (6th Edition)

$\dfrac{1}{16}$
(i) $a^m \cdot a^n = a^{m+n}$ (ii) $\dfrac{a^m}{a^n}=a^{m-n}$ (iii) $a^{-m} = \dfrac{1}{a^m}, a \ne 0, m \gt0$ Use rule (ii) to find: $=2^{3-7} \\=2^{-4}$ Use rule (iii) to find: $=\dfrac{1}{2^4} \\=\dfrac{1}{16}$