Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises: 99

Answer

$\frac{81}{16t^{4}}$

Work Step by Step

According to the definition of negative exponents, $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$). Therefore, $(\frac{2t}{3})^{-4}=\frac{1}{(\frac{2t}{3})^{4}}=(\frac{3}{2t})^{4}=\frac{3^{4}}{2^{4}t^{4}}=\frac{81}{16t^{4}}$.
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