## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\dfrac{1}{7}$
Using $a^{-x}=\dfrac{1}{a^x}$, the given expression, $7^{-1}$, is equivalent to \begin{array}{l} \dfrac{1}{7^1} \\\\= \dfrac{1}{7} .\end{array}