Answer
$\dfrac{1}{16}$
Work Step by Step
(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:
$=\dfrac{1}{2^4}
\\=\dfrac{1}{16}$
