Subject:
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Re: On a scale of 0 to 10?
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Wed, 26 Mar 2003 22:19:34 GMT
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Viewed:
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398 times
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In lugnet.off-topic.geek, John Riley writes:
> 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
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 figured out that B has 2 roots: 0
and 2.62502392. How would your average math wiz solve for B above?
DaveE
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Message has 1 Reply:
 | | Re: On a scale of 0 to 10?
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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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Message is in Reply To:
| | 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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