Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 58

$$s=\frac{1}{5}(t-5)+\ln 5$$

Given $$s(t)=\ln t, \quad t=5$$ Since at $t=5$, $s(t) = \ln 5$ and $$s'(t)= \frac{1}{t} $$ Then $ m= s'(t)\bigg|_{t=5}=\dfrac{1}{5}$ Hence, the tangent line is given by \begin{align*} \frac{s-s_1}{t- t_1}&=m\\ \frac{s-\ln 5}{t- 5}&=\dfrac{1}{5} \\ s&=\frac{1}{5}(t-5)+\ln 5 \end{align*}