College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Concept and Vocabulary Check: 14


$2$, $8$

Work Step by Step

As demonstrated on pages 45 and 46; $16^{\frac{3}{4}} = (\sqrt [4] {16})^{3} = (2)^{3} = 8$
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