College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.4 - Rational Exponents and Radicals - P.4 Exercises - Page 30: 71



Work Step by Step

The exponent of $x$ (which is 5) is lower then the index (9). This means that the radical is already in simplest form. RECALL: $\sqrt[n]{a^m} = a^{\frac{m}{n}}$ Thus, $\sqrt[9]{x^5} = x^{\frac{5}{9}}$
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.