Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 7 Exponential and Logarithmic Functions - Mixed Review of Problem Solving - Lessons 7.5-7.7 - Page 537: 5b

Answer

See below

Work Step by Step

We have the formula $N=\ln(E+1)$ Plug in values: $0.1=\ln(E+1)\\e^{0.1}=E+1\\E=e^{0.1-1}$
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