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I'm actually a programmer so I'm sorry if this is a stupid question.

I am trying to solve for $x$, but I got stuck at this step:

$e^{\frac{q1 + x}{b}} = e^{\frac{p - x\cdot o}{mb}} - e^{\frac{q2}{b}}$

I'm not sure how to get x by itself because it's in the exponent in both places if I do: $\ln(e^{\cdots})$

I'll end up with only one $x$ being solved for :( ie.

$x = b\ln(e^{\frac{p - x\cdot o}{mb}} - e^{\frac{q2}{b}}) - q1$

obviously that doesn't get me anywhere

I considered that there might be 2 (or more) answers to this problem (maybe one positive one negative?), but $x$ is guaranteed to be a positive number.

maybe I need to represent that in the problem somehow?

edit: more info

all values are known except for $x$; they are plugged into the equation by a computer program.

If it is possible to find x with arbitrary numbers plugged in for all the other values I'd like to see how to do that (I didn't see a way to solve it that way)

edit 2: realistic values

b = 5000 o = -2495 m = 3000 q1 = 90 q2 = 105 p = 75 

$e^{\frac{90 + x}{5000}} = e^{\frac{75 - x(-2495)}{3000\cdot 5000}} - e^{\frac{105}{5000}}$

reduced...

$e^{\frac{90 + x}{5000}} = e^{\frac{75 + 2495x}{15000000}} - e^{0.021}$

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    No closed form solution. And (I suspect) no solution for those values.2011-01-21

1 Answers 1

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Set $y = e^{x/15000000}$

This gives us

$ Ay^{3000} = By^{2495} - C$

Which is a polynomial, and I suppose can be solved easily using standard numerical methods.

Hope that helps.

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    And if we require x to be positive, we look for solutions of the polynomial $\gt 1$.2011-01-21