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: 5c

Answer

See below

Work Step by Step

From part a, $E=e^{0.05}-1$ From part b, $E=e^{0.1}-1$ Obtain: $\frac{e^{0.1}-1}{e^{0.05}-1}\\=\frac{(e^{0.05})^2-1}{e^{0.05}-1}\\=\frac{(e^{0.05}-1)(e^{0.05}+1)}{e^{0.05}-1}\\=e^{0.05}+1$
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