Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - Review Exercises - Page 52: 22

Answer

$k=\dfrac{\ln 8}{4}$

Work Step by Step

We are given the equation: $48=6e^{4k}$ Divide both sides by 6: $\dfrac{48}{6}=\dfrac{6e^{4k}}{6}$ $8=e^{4k}$ Apply logarithm: $\ln 8=\ln e^{4k}$ $\ln 8=4k\ln e$ $\ln 8=4k$ Divide both sides by 4: $\dfrac{\ln 8}{4}=\dfrac{4k}{4}$ $k=\dfrac{\ln 8}{4}$
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