Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 4 - Polynomials - 4.2 Negative Exponents and Scientific Notations - 4.2 Exercise Set: 129

Answer

$\dfrac{1}{2^{12}}$

Work Step by Step

$8^{-3} \cdot 32 \div 16^2 \\=(2^3)^{-3} \cdot 2^5 \div (2^4)^2$ Use the power rule $(a^m)^n=a^{mn}$ to obtain: $=2^{3(-3)} \cdot 2^5 \div 2^{4(2)} \\=2^{-9} \cdot 2^5 \div 2^8$ Use the rule $a^m \cdot a^n = a^{m+n}$ to obtain: $=2^{-9+5} \div 2^8 \\=2^{-4} \div 2^8$ Use the rule $a^m \div a^n = a^{m-n}$ to obtain: $=2^{-4-8} \\=2^{-12}$ Use the negative exponent rule ($a^{-m} =\frac{1}{a^m}, a\ne 0$) to obtain: $=\dfrac{1}{2^{12}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.