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: 69

Answer

$=\log(\frac{x(x-1)}{7})$

Work Step by Step

$= \log x + \log (x^{2}-1) - \log 7 - \log (x+1)$ $= \log(x(x^{2}-1)) - \log 7 - \log (x+1)$ $= \log(\frac{x(x^{2}-1)}{7}-\log(x+1))$ $=\log (\frac{x(x^{2}-1)}{7(x+1)})$ $=\log (\frac{x(x+1)(x-1)}{7(x+1)})$ $=\log(\frac{x(x-1)}{7})$
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