Is it possible to linearize $y = axe^{bx}$ in a linear regression?

Question

Is it possible to linearize $y = axe^{bx}$ in a linear regression?

My try: $y = axe^{bx} \to ln(y/x) = ln(a)+bx$.

Can I do $Y = ln(y/x)$. $A =ln(a)$ and $B = b$?

Thanks.

Answer 1

It could be some misunderstanding somewhere (probably a wording problem).

Let us suppose that you have data points $(x_i,y_i)$ and you want to fit the model $$y = axe^{bx}$$ Taking logarithms $$\log(y)=\log(a)+\log(x)+b x=\alpha+\log(x)+b x$$ So, the residue is $$r_i=\alpha+\log(x_i)+b x_i-\log(y_i)=\alpha+b x_i-\log(\frac{y_i}{x_i})$$ which is the same as for model $$z=a e^{bx}\implies \log(z)=\alpha+b x$$ (using $z=\frac y x$).

So, the original model can be linearized. A simple linear regression will give $\alpha$ (then $a=e^\alpha$) and $b$. But, for sure, this is just the preliminary step before a nonlinear regression since what is measured is $y$ and not $\log(y)$ and still less $\log(\frac y x)$.