College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.5 - Properties of Logarithms - 6.5 Assess Your Understanding: 92

Answer

$y=Ce^{-2x}$

Work Step by Step

Subtract $\ln{C}$ on both sides: $\ln{y} - \ln{C}=-2x$ RECALL: For positive real numbers M and N: $\ln{M} - \ln{N} = \ln{\left(\dfrac{M}{N}\right)}$ Use the rule above to obtain: $\ln{\left(\dfrac{y}{C}\right)}=-2x$ RECALL: $\ln{M}= y \longrightarrow e^y=M$ Use the rule above to obtain: $e^{-2x}=\dfrac{y}{C}$ Multiply by $C$ on both sides of the equation to obtain: $C \cdot e^{-2x} = C \cdot \dfrac{y}{C} \\Ce^{-2x} = y \\y=Ce^{-2x}$
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