Recent content by totti

exp(ax) = bx + c; I think this might be the solution: let bx + c = y/a; a*x = (y - c*a)/b; exp(y/b)*exp(-c*a/b) = y/a; or -(a/b)*exp(-c*a/b) = (-y/b)*exp(-y/b); or -y/b = W(-(a/b)*exp(-c*a/b)); or y = -b*W(-(a/b)*exp(-c*a/b)); where y = a*(bx+c); You can check for...

I don't know if the solution is accurate or not but here is the method. I have used a couple of approximations and assumptions. Assuming a > 0 and for x << a we can rewrite the equation as exp(a-x) = (x-c)/(b-x) ; adding one two both sides : 1 + exp(a-x) = (b-c)/(b-x); and...