I want to fit data to an exponential model, but the table given to me in the exercise contains a $0$ as a value for the independent variable, which is making the argument of the exponential indeterminate. My model is $y=ae^{b/x}$. The goal is to linearize this and apply linear regression.
Since the value of $y(0)$ in the data table is finite, I deduced that $b$ should be negative (the values of $x$ in the table are increasing, so I'm taking the limit from the right), and that $a=y(0)$.
My problem is with the linearization step.
- Option 1: Should I just disregard $(0,\,y(0))$ as a data point and proceed to use $\ln y=\ln a+b(1/x)$,
- Option 2: or should I use it as mentioned above to deduce $a$, and fit the data to $\underbrace{(1/b)}_{\text{slope}}x+\underbrace{0}_{\text{intercept}}=\frac1{\ln(y/a)}$?
My issue with the first option is obviously letting go of data. As for the second option, the indeterminacy is still present for $y=a$, and, should I again disregard the first data point, I doubt that the formula for the intercept will give me $0$, so that I will have to decide on which value to use.
As a third option, I thought of using a value like $10^{-6}$ instead of $0$, since we're doing approximations anyhow.
Any tips? Thank you for your time!
$$y=a\:e^{b/x}$$ $$\ln(y)=\ln(a)+\frac{b}{x}$$ $$x\ln(y)=\ln(a)\:x+b$$ Let : $$Y=x\ln(y)\quad\text{and}\quad c=\ln(a)$$ Compute the data $Y(x)=x\ln\left(y(x)\right)$.
Try the linear regression for $b$ and $c$ : $$Y=b+c\:x$$