Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 1 - Ingredients of Change: Functions and Limits - 1.1 Activities - Page 10: 22

Answer

(a) $t= -4.904$ (b) $t= -12.951$

Work Step by Step

(a) Expression given: $ \frac{3}{8e^{0.2t}}+5 = 6$ Multiply both sides by $8e^{0.2t}$: $3+5\times(8e^{0.2t}) = 6\times(8e^{0.2t})$ Expand: $ 3+40e^{0.2t} = 48e^{0.2t}$ Rearrange: $ 3 = 8e^{0.2t}$ Simplify: $ \frac{3}{8} = e^{0.2t}$ Take the natural logarithm of both sides: $\ln \frac{3}{8} = 0.2t$ Solve using calculator or quadratic formula to find: $ t = \frac{\ln \frac{3}{8}}{0.2} \approx -4.904$ (b) Expression given: $ \frac{3}{8e^{0.2t}}+5 = 10$ Multiply both sides by $8e^{0.2t}$: $3+5\times(8e^{0.2t}) = 10\times(8e^{0.2t})$ Expand: $ 3+40e^{0.2t} = 80e^{0.2t}$ Rearrange: $ 3 = 40e^{0.2t}$ Simplify: $ \frac{3}{40} = e^{0.2t}$ Take the natural logarithm of both sides: $\ln \frac{3}{40} = 0.2t$ Solve using calculator or quadratic formula to find: $ t = \frac{\ln \frac{3}{40}}{0.2} \approx -12.951$
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