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

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

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*}