Solve y=1- A(1-e^(-kx)) + mx +b

[O_chemist]

At work today one of the biologist came and asked me with help on solving a problem. I need a little hint as I still would like to solve this problem.

y=1- A(1-e^(-kx)) + mx +b

A, k, m and b are all constants

I need to solve for x but it results in getting a -kx-ln(x) = some junk
From what I can see there isn't a way to do this analytically as I will go in an endless cycle of e's and ln's. I did some research about using iterative method for solving this. What I want to know is from people who do this more often is there a more simple method I could use? Just give me a hint ;) .

Regards,
Caleb

 

[Mark44]

At work today one of the biologist came and asked me with help on solving a problem. I need a little hint as I still would like to solve this problem.

y=1- A(1-e^(-kx)) + mx +b

A, k, m and b are all constants

I need to solve for x but it results in getting a -kx-ln(x) = some junk
From what I can see there isn't a way to do this analytically as I will go in an endless cycle of e's and ln's. I did some research about using iterative method for solving this. What I want to know is from people who do this more often is there a more simple method I could use? Just give me a hint ;) .

There's the Lambert W function, https://en.wikipedia.org/wiki/Lambert_W_function, but the way I would approach it would be as a numerical solution.
Your equation is equivalent to $1 - A(1 - e^{-kx}) + mx + b - y = 0$, with A, k, m, and b known constants, and the value for y known.

Pick a value for x, and plug it in. Is the result zero? Probably not, but if so, you've found the solution.
If not, pick a different value for x, and evaluate the expression. Choose values for x that make the expression on the left side as close to zero as you can.