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: 8

Answer

$\frac{27}{a^{12}b^{21}}$

Work Step by Step

We are given the expression $(\frac{3a^{-4}}{b^{7}})^{3}$. We can use the power of a quotient rule to simplify, which holds that $(\frac{a}{b})^{n}=\frac{a^{n}}{b^{n}}$, $b\ne0$ (where a and b are real numbers, and n is an integer). $(\frac{3a^{-4}}{b^{7}})^{3}=\frac{(3a^{-4})^{3}}{(b^{7})^{3}}$ To simplify further, we can use the power rule, which holds that $(a^{m})^{n}=a^{m\times n}$ (where a is a real number, and m and n are integers). $\frac{(3a^{-4})^{3}}{(b^{7})^{3}}=\frac{3^{3}a^{-4\times3}}{b^{7\times3}}=\frac{27a^{-12}}{b^{21}}=\frac{27}{a^{12}b^{21}}$
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.