Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 59


$$s=-(t-1)+\ln 4$$

Work Step by Step

Given $$s(t)=\ln (8-4 t) \quad t=1$$ Since at $t=1$, $s(t) = \ln 4$ and $$s'(t)= \frac{-4}{8-4t} $$ Then $ m= s'(t)\bigg|_{t=1}=-1$ Hence, the tangent line is given by \begin{align*} \frac{s-s_1}{t- t_1}&=m\\ \frac{s-\ln 4}{t- 1}&=-1 \\ s&=-(t-1)+\ln 4 \end{align*}
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