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Is there an algebraic solution for the to find the intersection of the following two functions for values of $x\geq 0$:
$$f_1(x)=1-2e^{-x/a}=f_2(x)=-1+2e^{-x/b}$$
$a$ and $b$ are positive constants.

The equation can be simplified to:
$$e^{-x/a}+e^{-x/b}=1$$

A Plot is here:

http://img194.imageshack.us/img194/8276/inversionrecovery.jpg

I am searching for the $x$-value of the intersection in the second plot (this is for an inversion recovery experiment inf magnetic resonance).

If there is no algebraic solution, can you suggest a numerical algorithm for this problem?

Thanks

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    In some special cases we can solve. For example $a=1,b=2$ ... this amounts to a quadratic equation followed by a logarithm.2011-11-24
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    On another hand if $\frac{1}{a}$ and $\frac{1}{b}$ are algebraic numbers which are linearly independent over $Q$, then $x$ has to be transcendental... So depending by what "algebraic" means, in general I would expect the answer to be no....2011-11-24
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    Let $e^{-x/{(a*b)}}=t$, then the equation can be simpified to be :$t^b+t^a=1$. The group theroy tells us that there is no general formula to solve polynomial equations whose oder is greater than 4. And I don't think we can find a general formula to solve this equation, for any a and b.(I am not professional, just my thought...) A possible solution is the numerical solution.2011-11-24
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    So far as I remember, sums of exponentials aren't that easy to solve (barring things like the hyperbolic functions). I too doubt that you'd get a symbolic solution. Be happy with Newton-Raphson.2011-11-24
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    Well, $x^6+x=1$ has no solution in radicals, although the solution is "algebraic" of course.2011-11-24

2 Answers 2

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I agree with the suggestions in the comments that for most values of $a,b$ there will be no algebraic solution. As for a numerical algorithm, are you familiar with Newton's Method (often called Newton-Raphson)?

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    Thanks, I feared that there is no algebraic solution, so I will go with Newton.2011-11-24
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Why not reduce to $y^{b/a}+y=1$, then you have only a one-parameter family to solve.

graph

The solution for $y^r+y=1$, expanded in a Taylor series near $r=1$ is
$$ \frac{1}{2} + \frac{\operatorname{ln} (2)}{4}(r - 1) - \frac{\operatorname{ln} (2)}{8} (r - 1)^{2} - \frac{\operatorname{ln} (2) \bigl(-6 + 2 \operatorname{ln} (2)^{2} - 3 \operatorname{ln} (2)\bigr)}{96} (r - 1)^{3} + \frac{\operatorname{ln} (2) \bigl(2 \operatorname{ln} (2) + 1\bigr) \bigl(\operatorname{ln} (2) - 2\bigr)}{64} (r - 1)^{4} + \operatorname{O} \bigl((r - 1)^{5}\bigr) $$