Simplify $\log_e(1-e^{a+bt})$

Question

I have to simplify the following equation in which 't' is the independent variable while 'a' and 'b' are coefficients or constants.

$\log_e(1-e^{a+bt})$

I tried the reduction for-

$\log(a-b) = \log(a(1 - b/a))$

But that returns back to the original form. Any suggestions?

Thanks!

I am working on a problem which involves finding a best fit for curve

$m = m_0*(1-e^{a+b*t})$

according to given data points for "m" and "t" by estimating "a" and "b" since $m_0$ is given.

In order to do this, I need to find a transformation to reduce the problem to a regression line.

I was trying to transform the equation by taking the log to base e of both sides but then I am stuck with the simplification as mentioned above.

Answer 1

If you have $n$ data points $(t_i,y_i)$ and you want to fit the model $$y=\log(1-e^{a+bt})$$ just rewrite it as $$e^y=1-e^{a+bt}\implies 1-e^y=e^{a+bt}\implies \log(1-e^y)=a+bt$$ So, defining first, for each data point, a new variable $z_i=\log(1-e^{y_i})$, a linear regression $z=a+b t$ will give you good estimates of parameters $a$ and $b$.

When this is done, using theses guesses, you should safely start a nonlinear regression.