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I am trying to solve for [Lambda] as a function of a (unknown) from this expression

202.[2.51521 + 1/(-1 + E^(202. λ))] + 
 802.[2.52457 + 1/(-1 + E^(802. λ))] + 
 1802.[2.52632 + 1/(-1 + E^(1802. λ))] + 
 3202.[2.52694 + 1/(-1 + E^(3202. λ))] + 
 5002.[2.52722 + 1/(-1 + E^(5002. λ))]=a

I tried with Reduce and FindRoot, and as neither of them helped I think closed form expression can not be obtained. Any suggestion how to get numerical solution? Thanks in advance.

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  • $\begingroup$ You can use FindRoot to solve once given specific values for a. $\endgroup$ – Daniel Lichtblau Apr 23 '15 at 17:20
  • $\begingroup$ @DanielLichtblau Only after those square brackets are replaced by parenthesis $\endgroup$ – Dr. belisarius Apr 23 '15 at 17:21
  • $\begingroup$ Anyway, the function is too steep and you'll probably need a few tricks $\endgroup$ – Dr. belisarius Apr 23 '15 at 17:22
  • $\begingroup$ @DanielLichtblau Thanks. But I want to express λ as a function of a. Sorry about the square brackets. $\endgroup$ – reach2brb Apr 23 '15 at 17:41
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    $\begingroup$ [I have nothing against square brackets.] $\endgroup$ – Daniel Lichtblau Apr 23 '15 at 19:23
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In such cases I prefer to get an idea from numerical solution.

f[x_] = 202. (2.51521 + 1/(-1 + E^(202. x))) + 802. (2.52457 + 1/(-1 + E^(802. x))) + 1802. (2.52632 + 1/(-1 + E^(1802. x))) + 3202. (2.52694 + 1/(-1 + E^(3202. x))) + 5002. (2.52722 + 1/(-1 + E^(5002. x)));
data = Table[{a, x /. FindRoot[f[x] == a, {x, -0.001}]}, {a, -100, 100, 10}];
ListPlot[data]
Fit[data, {1, a, a^2}, a]

enter image description here

$-0.000231828 - 1.11206 \times 10^{-8} a - 5.50983 \times 10^{-13} a^2$

So we have three steps here. First I solve it for different a, then store them in data. And then I go for a polynomial fit. As you can see within the defined range of a [-100,100] it is almost linear. You can check with different a values. For higher a you have to take more terms in your polynomial.

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