College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Review Exercises: 35

Answer

$i$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ i^{-27} ,$ 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} i^{-28+1} \\\\= i^{-28}\cdot i^1 \\\\= i^{-28}\cdot i .\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{1}{i^{28}}\cdot i .\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})^{14}}\cdot i .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{(-1)^{14}}\cdot i \\\\= \dfrac{1}{1}\cdot i \\\\= 1\cdot i \\\\= i .\end{array}
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