## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 4 - Review - Exercises: 74

#### Answer

$k=-\dfrac{\ln10000}{15}\approx-0.614023$

#### Work Step by Step

$e^{-15k}=10000$ Apply $\ln$ to both sides of the equation: $\ln e^{-15k}=\ln10000$ Take $-15k$ down to multiply in front of its $\ln$: $-15k\ln e=\ln10000$ Since $\ln e=1$, this equation becomes: $-15k=\ln10000$ Solve for $k$: $k=-\dfrac{\ln10000}{15}\approx-0.614023$

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