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...