College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3: 39

Answer

$= 1 + 2\log x + \frac{1}{3}\log(1-x) - \log7-2\log(x+1)$

Work Step by Step

$= \log[\frac{10x^{2}\sqrt[3] {1-x}}{7(x+1)^{2}}]$ $= \log(10x^{2}\sqrt[3] {1-x}) - \log(7(x+1)^{2})$ $= \log(10x^{2}(1-x)^{\frac{1}{3}}) - \log(7)-\log((x+1)^{2})$ $=\log(10x^{2})+\log(1-x)^{\frac{1}{3}} - \log7-2\log(x+1)$ $=\log(10) + \log x^{2} + \frac{1}{3}\log(1-x) - \log7-2\log(x+1)$ $= 1 + 2\log x + \frac{1}{3}\log(1-x) - \log7-2\log(x+1)$
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.