Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Cumulative Review - Page 408: 32b

Answer

$y^{10b+3}$

Work Step by Step

Using the laws of exponents, the given expression, $ \dfrac{\left( y^{4b} \right)^3}{y^{2b-3}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{y^{4b(3)}}{y^{2b-3}} \\\\= \dfrac{y^{12b}}{y^{2b-3}} \\\\= y^{12b-(2b-3)} \\\\= y^{12b-2b+3} \\\\= y^{10b+3} .\end{array}
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