Answer
$\log \dfrac {1}{x^{2}\sqrt {x}}=-\log x^{2}\sqrt {x}$
Work Step by Step
$\log _{3}\sqrt {x}-\log _{3}x^{3}=\log _{3}\dfrac {\sqrt {x}}{x^{3}}=\log \dfrac {1}{x^{2}\sqrt {x}}=-\log_{3} (\frac{1}{x^{5/2}})$
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 6 - Section 6.5 - Properties of Logarithms - 6.5 Assess Your Understanding - Page 459: 59
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
