Cómo resolver para x cuando está en el exponente de 2 variables

Question

En realidad, soy un programador así que, lo siento si es una pregunta estúpida.

Estoy tratando de resolver para $x$, pero me atoré en este paso:

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

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...

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

Answer 1

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

Esto nos da

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

Que es un polinomio, y supongo que se puede resolver fácilmente utilizando el estándar de métodos numéricos.

Espero que ayude.