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(...) Thought about that... If necessary I guess, but I'm pretty sure there's a way to do it. I think what I need is: s = Ae^(q + B) + C And at the moment, I've solved: C = -Ae^(B - 1) A = 10 / (e^(B + .25) - e^(B - 1)) And I'm scratching my head (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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| | Re: On a scale of 0 to 10?
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(...) *snip* (...) summed exponents like that can be re-written in this way: e^B * e^.25 + e^B * e^-1 = 2e^B (e^.25 + e^-1) * e^B = 2e^B Uh oh. Divide both sides by e^B and we've got (e^.25 + e^-1) = 2 1.652 = 2 Which ain't right. I think your math (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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(...) Oh yeah! (...) Oh... yeah. (...) Hm. Double checked the math-- it appears solid [1], which would mean that there's no viable solution for: s = Ae^(q + B) + C for coordinates (-1,0), (0,5), (0.25,10). Darn. Hm. I guess that also means that any (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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| | Re: On a scale of 0 to 10?
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Hi all. The reason the equation is unsolvable is because B is an unneeded constant. Why? A*e^(q+B)=A*(e^q)*(e^B) It is impossible to distinquish between A & e^B. What you need to solve is an equation of the form s=A*exp(B*q) + C This is the form of (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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(...) aHA! Much oblige! I now get (approx): s = 5.39 * e^(2.625q) - 0.39 Whew. Of course, now here's a totally different question. In order to get that point, I cheated. I couldn't solve: e^(B/4) + e^(-B) = 2 using algebra, but using other means, I (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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(...) I don't see an easy way to do this. I tried substituting x=e^B Then I get a polynomial of degree 5. Only the positive roots lead to real values of B. (since e^any real number > 0) This is solvable, but very difficult to do by hand, especially (...) (22 years ago, 28-Mar-03, to lugnet.off-topic.geek)
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