Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - Chapter Review - Review Exercises - Page 113: 17



Work Step by Step

Comparing the graphs, As x grows greater, for $a >0,$ $e^{x+a} =e^{x}\cdot e^{a}$ is greater than $e^{x}$ much more than $\ln(x+a)$ is greater than $\ln x$ The statement is true
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