## Thinking Mathematically (6th Edition)

$2$
(i) $a^m \cdot a^n = a^{m+n}$ (ii) $\dfrac{a^m}{a^n}=a^{m-n}$ (iii) $a^{-m} = \dfrac{1}{a^m}, a \ne 0, m \gt0$ (iv) $(a^m)^n=a^{mn}$ Use the applicable rules above to find: $\require{cancel}=\dfrac{20x^8}{10x^8} \\=\dfrac{20\cancel{x^8}}{10\cancel{x^8}} \\=\dfrac{20}{10} \\=2$