assessing linear relationships as logarithms I am teaching myself maths. I am not sure how to approach this problem. It is assessing linear relationships of the form $y=mx+c$ as logarithms. 
Here I have gotten as far as taking the gradient ($\log e$) of $\log \frac{s}{t} = -0.6363...$ so $e = 0.23$ but I do not know how to derive or separate $-n$ or what it represents. Could someone tell me how to proceed? Thank you.
Alan


We have
\begin{align}
s=ke^{-nt}\tag1
\end{align}
Taking logarithm both sides of $(1)$ yields
\begin{align}
\ln s&=\ln\left(ke^{-nt}\right)\\
\ln s&=\ln k +\ln\left(e^{-nt}\right)\qquad&\rightarrow\qquad\ln(ab)=\ln a+\ln b\\
\ln s&=\ln k-nt\ln e\qquad&\rightarrow\qquad\ln a^b=b\ln a\\
\ln s&=\ln k-nt\qquad&\rightarrow\qquad\ln e=1\\
\ln s&=-nt+\ln k\tag2
\end{align}
Let $y=\ln s$, $m=-n$, $x=t$, and $c=\ln k$, then use linear interpolation to obtain $m$ and $c$. We will get
$$
n=-m
$$
and
$$
k=e^c
$$
