Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.2 - More Work with Exponents and Scientific Notation - Exercise Set - Page 268: 63

Answer

$y^{15a+3}$

Work Step by Step

Using laws of exponents, the given expression simplifies to \begin{align*} & \dfrac{\left( y^{2a} \right)^8}{y^{a-3}} \\\\&= \dfrac{y^{2a(8)}}{y^{a-3}} \\\\&= \dfrac{y^{16a}}{y^{a-3}} \\\\&= y^{16a-(a-3)} \\\\&= y^{16a-a+3} \\\\&= y^{15a+3} .\end{align*}
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