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

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