Fitting y = a + exp(bt) to two points

Question

I would like to fit y = a + exp(bt) to the points (1,1) and (2,5). So I take the natural logarithm of the equation and apply it to both points:

ln(y1) = ln(a) + bt1
ln(y2) = ln(a) + bt2

With (1,1) = (t1, y1) and (2,5) = (t2, y2) so that:

ln(1) = ln(a) + b
ln(5) = ln(a) + 2b

Subtracting yields b = ln(5) = 1.609. I plug this back into my linearized system and I get ln(a) = -1.609, therefore a = 0.2. So far, so good. Do these numbers work in the original equations?

a = y1 - exp(bt1) --> 0.2 = 1 - exp(1.609) = 1 - 5 = -4 WRONG!
a = y2 - exp(bt2) --> 0.2 = 5 - exp(2 * 1.609) = 5 - 25 = -20 ALSO WRONG!

I would have thought that even though the initial equations were nonlinear, taking logarithms would have changed them to two linear equations in 2 variables {ln(a), b}. This should in theory produce a unique answer for the pair {a,b}.

Where am I going wrong?

Roberto

This might be better posted on http://math.stackexchange.com/ ? — Tom Lord, Jun 24 '16 at 15:57
I'm voting to close this question as off-topic because it is about [math.se] instead of programming or software development. — Pang, Aug 10 '16 at 03:18

score 1 · Answer 1 · answered Jun 24 '16 at 15:43

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Algebra is incorrect.

y = a + exp(bt)

Subtract a from both sides:

(y-a) = exp(bt)

Take the natural log of both sides:

ln(y-a) = bt

answered Jun 24 '16 at 15:43

duffymo

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Of course! So simple, but I was stuck in a mental rut. Thanks, duffymo. – R. Estevan Jun 26 '16 at 06:31

score 1 · Answer 2 · answered Jun 24 '16 at 15:47

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Log(x + y) Does not equal Log(x) + Log(y)

Your initial implementation of ln is wrong.

@duffymo has a good implementation of log.

answered Jun 24 '16 at 15:47

Nicholas Patton

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Fitting y = a + exp(bt) to two points

2 Answers2