assessing linear relationships as logarithms

Question

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

Answer 1

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 $$ k=e^c $$

Answer 2

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 $c$.