Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.5 Properties of Logarithms - 5.5 Assess Your Understanding - Page 306: 92

Answer

$y=Ce^{-2x}$.

Work Step by Step

We know that if $e^x=y$, then $\ln {y}=x$, hence because $-2x=\ln{(e^{-2x})}$, $\ln{y}=\ln{e^{-2x}} + \ln{C}$. We know that $\log_a {x}+\log_a {y}=\log_a {(x\cdot y)}$, hence $\ln{(y)}=\ln{(e^{-2x})}+\ln{(C)}=\ln{(Ce^{-2x})}.$ The base is same on the 2 sides on the equation (and it is not $1$), hence they will be equal if the exponents are equal.($\log_a{M}=\log_a{N} \longrightarrow M=N$.) Hence $y=Ce^{-2x}$.
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