$$y_i=\frac{(x_{0i}\beta_0+x_{1i})+\log(\beta_1^2 x_{2i})}{x_{3i}}+e_i\quad,\,i=1,\ldots,n,\,x_{pj}>0$$
I am wondering if this can be written as a linear model, I didn't think so because of the Beta1 term in the log part of the equation. But I would love some clarification.
Thanks!
Yes it can.
The subscript "$i$" is superfluous, so let's drop it. Assuming your parameters are the betas and the x's are the data, your model is
$$\begin{aligned} y &= \frac{x_0\beta_0 + x_1 + \log\left(\beta_1^2 x_2\right)}{x_3} + e \\ &= \beta_0 \frac{x_0}{x_3} + \frac{x_1}{x_3} + \log(\beta_1^2)\frac{1}{x_3} + \frac{\log(x_2)}{x_3} + e. \end{aligned}$$
It will be convenient to put this in a standard form with all offsets (constant terms) on the left, thus:
$$\begin{aligned} w &= y - \frac{x_1}{x_3} - \frac{\log(x_2)}{x_3} \\&= \beta_0 \frac{x_0}{x_3} + \log(\beta_1^2)\frac{1}{x_3} + e \\ &= \gamma_0 z_0 + \gamma_1 z_1 + e \end{aligned}$$
where $\gamma_0=\beta_0$ and $\gamma_1 = \log \beta_1^2$ are parameters and $w,$ $z_0 = x_0/x_3,$ and $z_1=1/x_3$ are re-expressions of the data values. These will all be well-defined (that is, $x_3$ will not be zero and $\beta_1$ will not be zero) if and only if the original model makes any sense.
This re-expressed model obviously is linear in both the new parameters and the re-expressed data, as well as in the error term $e:$ this makes it a plain vanilla multiple regression model.
Finally, the original parameters (the betas) can be recovered from the estimated parameters $\hat\gamma_i$ as
$$\hat\beta_0 = \hat\gamma_0;\ \hat\beta_1 = \pm\exp(\hat\gamma_1/2).$$
How to tell the difference between linear and non-linear regression models?, URL (version: 2016-03-18): https://stats.stackexchange.com/q/148713
