Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Mid-Chapter Check Point - Page 477: 22

Answer

$$19+20 \ln x$$

Work Step by Step

We simplify as follows: $$\ln (e^{19}x^{20})= \ln e^{19} + \ln x^{20}= 19 \ln e + 20 \ln x = 19 + 20 \ln x$$
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