How to interpret b in $y=x^{e^{bz}}$ in nonlinear regression?

Question

What is the correct way to interpret b in this nonlinear equation $y=x^{e^{bz}}$?

I've estimated the model and b seems to be the percent change in y with a unit change in z, but I am unsure how to show this.

The derivative of y w.r.t z seems to be:

• $dy/dz=bx^{e^{bz}}ln(x)e^{bz}$ then
• $dy/dz=byln(x)e^{bz}$

A one unit change in z would lead to:

• $=byln(x)e^{b(z+1)}$
• $=byln(x)e^{bz}e^b$

It's at this point I get stuck. I feel like there is some simplifying to get $dy/dz=ye^b$ that I'm not noticing.

Is this even the right approach?

For example, it's easy to show that for $y=Ae^{bx}$ that y increases at a constant relative rate of b.

• $=Ae^{b(x+1)}$
• $=Ae^{bx+b}$ -> $Ae^{bx}e^b$
• $=ye^b$

Answer 1

$$\begin{align} y = & ~x^{e^{bx}}\\ \log(y) = & ~\log(x) e^{bx}\\ \log(\log(y)) = & ~bx + \log(\log(x)) \\ \log\left(\frac{\log(y)}{\log(x)}\right) = & ~bx \\ \end{align}$$

If you set $Z = \log\left(\frac{\log(y)}{\log(x)}\right)$ then you can write your equation as follows:

$$Z = bx,$$

which is suitable for regression purposes (i.e. estimate $b$ given data $x$, $y$ and hence $Z$).

Note that this approach only works when $\frac{\log(y)}{\log(x)} > 0.$