TITLE: Why can we linearize the exponential regression?
QUESTION [10 upvotes]: When we are using observed data $(x_1,y_1)\ldots(x_m,y_m)$ for the exponential model $$y(x)=a_1\mathrm{e}^{a_2x}~(a_1>0),$$ it is natural to think about the linearized model $$\ln y=\ln a_1+a_2x.$$
It is not hard to understand this approach, but my question is, how can we prove rigorously that we are obtaining the same results? I.e., the following two optimization problems for $a_1$ and $a_2$
$$\text{minimize}\sum_{k=1}^m(a_1\mathrm{e}^{a_2 x_k}-y_k)^2$$ and $$\text{minimize}\sum_{k=1}^m(\ln a_1+a_2 x_k-\ln y_k)^2$$ yield the same $a_1$, $a_2$?
My attempts using multivariate calculus failed, and I can only say that since the problems are "equivalent" with unique solution, the answer should be unique. Is that acceptable?

REPLY [2 votes]: You can try the following.
Suppose your data has just two points $(x_1,y_1)=(1,1), (x_2,y_2) = (2,2)$. Now fix one of the parameters in both of the problems, say $ a_2 = 1 $. Use calculus to determine what the corresponding 'minimizing' $ a_1 $ in each of the problem will be. Do they match? This should be easy to check due to the small size of data.