linear equations and logarithms: variables s, t related by equation s = ke^-nt

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alan82

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I have a question in my A level text book that reads: Two variables s and t are related by a law of the form s=ke^-nt. The values in the table were obtained from an experiment. Show graphically that these values do verify the relationship and use the graph to find approximate values of k and n.

t11.522.53
s123059026014060

in this format the relationship is not linear so we use logs but im not quite sure how to approach it. to begin with do i convert it to: log s = -nt log k + log e and tabulate -t and log s? or do i need to split the -n and t somehow?
 
First off, I suspect your equation is really meant to be s = ke^(-nt). Assuming that's the case, be sure you understand why those grouping symbols are very important. That small thing aside, one way to begin would be to simply use the information to create five equations:

\(\displaystyle k \cdot e^{-n} = 1230\)

\(\displaystyle k \cdot e^{-1.5n} = 590\)

\(\displaystyle k \cdot e^{-2n} = 260\)

etc.

Each of these equations looks very similar and shares a lot of common terms. For instance, the third (t = 2) can be rewritten as:

\(\displaystyle k \cdot \left( e^{-n} \right)^2 = 260\)

Where does that lead you?
 
the exercises im doing have come after a short section on using logarithms to extract linear relationships from non linear ones. so aside from taking logs of the equations so that they are similar to the form y=mx+c, im not sure, and i dont know how to express (e-n)t ​as a logarithm
 
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alan82 said:
I have a question in my A level text book that reads: Two variables s and t are related by a law of the form s=ke^-nt. The values in the table were obtained from an experiment. Show graphically that these values do verify the relationship and use the graph to find approximate values of k and n.

t11.522.53
s123059026014060

in this format the relationship is not linear so we use logs but im not quite sure how to approach it. to begin with do i convert it to: log s = -nt log k + log e and tabulate -t and log s? or do i need to split the -n and t somehow?
Click to expand...
Are you supposed to calculate the best-fit curve?

A logarithmic function does not go through those points (exactly).

A power curve gives you a better fit (but not exact).
 
no, you are supposed to convert it to a linear relationship by taking logs and draw a straight line graph.
 
here is an example of a previous question and how i solved it:


The table gives values for a set of related variables and a law that relates them. by drawing a straight line graph find approximate values for a and b

ay=b^x

x5678
y1.072.134.278.53


reducing to logs gives: log a + log y = x log b.
comparing with y=mx+c we have log y = x log b - log a where log b is the gradient and -log a is the y intercept

tabulating x against log y we have

x5678
log y0.0290.3280.630.931

which is a linear relationship as you will see if you graph it

taking the gradient we have 0.328-0.029/6-5 = 0.299
so log b = 0.299 --> b = roughly 2

to find a we use a point on the graph, lets say (6, 0.328), and comparing with

log y = x log b - log a and y=mx+c we have 0.328 = 0.299(6) + c --> c = -1.466 = - log a

a = -log a --> a = roughly 30
 
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i figured it out. i just didnt initially realise that e refered to the constant e and thought that it was some unknown
 
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