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

 A: 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
$$
k=e^c
$$
A: If you take logarithms of both sides of the formula mentioned in the problem you get:
$$\ln(s) = \ln(k)-n t$$
So in the terminology of your post $y=\ln(s), c=\ln(k), m=-n, x=t$.
To create the graph, augment the table with one more row and populate it with $\ln(s)$ taking logaritms of the second row. Then use the $t$-values of the first row for the horizontal coordinate the values of the third row as the vertical coordinate. You will have 5 points in your graph and if they are close to a straight line you have your graphical verification. Next calculate (graphicaly, an approximation) the slope of that line and you get $m$. Finaly the point that it intersects the vertical axis gives you $c$.
