Linearization before curve fitting

Question

I am currently studying at college curve fitting to a set of experimental data. One of our activities/homework was to fit the curve $y=ax^b$ to the set of points of the position of a free-falling object we filmed. My professor stated that it was always necessary to first linearize the data, by taking the logarithm of the $x$ and $y$ values, and then do a least squares regression on this new data, and that the result of this linear regression would be the final correct value. However, doesn't this change the final result compared to a non-linear regression? Not only that, wouldn't the result change depending on the units, because $\log(y)$ is the logarithm of a quantity which has units? If it changes the final result of the regression, which method is better to use in an experimental context such as the one mentioned above?

Answer 1

Your professor is totally correct. The model $$y=a \,x^b$$ is nonlinear because of $b$. So, in a first step, you linearize $$\log(y)=\log(a)+b\log(x)=c+b\log(x)$$ and a linear regression gives the estimates of $b$ and $a=e^c$.

Now, you start with them the nonlinear regression what you must do since what is measured is $y$ and not any of its possible transforms.

Edit

When you linearize the model, the residue at point $i$ is $$\text{res}_i=\log\big[y_i^{\text{(calc)}}\big]-\log\big[y_i^{\text{(exp)}}\big]$$ which can rewrite as $$\text{res}_i=\log\Bigg[\frac{y_i^{\text{(calc)}} } {y_i^{\text{(exp)}} } \Bigg]=\log\Bigg[1+\frac{y_i^{\text{(calc)}}-y_i^{\text{(exp)}} } {y_i^{\text{(exp)}} } \Bigg]$$ So, if the error is small $$\text{res}_i \sim \frac{y_i^{\text{(calc)}}-y_i^{\text{(exp)}} } {y_i^{\text{(exp)}} }$$ which is the relative error.

In practice, when the range of the $y_i^{\text{(exp)}}$ is very large, it is often preferable to stop at this point. This is the case for physical properties like vapor pressure.