## College Algebra (11th Edition)

$-i$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\dfrac{1}{i^{17}} ,$ use the laws of exponents and the equivalence $i^2=-1.$ $\bf{\text{Solution Details:}}$ Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{i^{18-1}} \\\\= \dfrac{1}{i^{18}\cdot i^{-1}} .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{i^{2\cdot9}\cdot i^{-1}} \\\\= \dfrac{1}{(i^2)^9\cdot i^{-1}} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{i}{(i^2)^9} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{i}{(-1)^9} \\\\= \dfrac{i}{-1} \\\\= -i .\end{array}