Fitting an exponential function

Question

Could someone please help me with this exercise and tell me if I am on the right track?

Assume that based on a data set with a large number of observations of an independent variable $x$ and a dependent variable $y$ we have estimated the coefficients $\alpha$ and $\beta$ in $\ln(y) = \alpha + \beta \cdot x$.

1. Compute the predicted change in $y$ (denoted $\Delta y$) that results from increasing $x$ by $\Delta x$ (from $x$ to $x + \Delta x$).

I thought maybe to put $e$ in the equation. $e(\ln(y))=e^\alpha + \beta (x+\Delta x)$ so $e(\ln)$ cancel and I have $y=e^\alpha + \beta(x+\Delta x)$ where I have substituted $x+\Delta x$ into $x$. Would this equation show the change in $y$?

2. When is $\Delta y$ positive/negative/zero?

If it is an exponential function, then it just approaches zero no?

Help appreciated! Thank you!

Answer 1

For $$ \ln(y) = \alpha + \beta \cdot x $$ We get $$ \frac{d \ln(y)}{dx} = \frac{1}{y} \frac{dy}{dx} = \beta $$ and notice $$ \Delta y \approx \frac{dy}{dx}\Delta x = \beta \, y \, \Delta x $$

Answer 2

You need to find $\Delta y$ in $$y(x+\Delta x)=y(x)+\Delta y=e^{\alpha + \beta (x+\Delta x)}=e^{\alpha+\beta x}e^{\beta \Delta x}=y(x)e^{\beta \Delta x}\implies \Delta y=y(x)\left(e^{\beta \Delta x}-1 \right)$$

This is the amount that $y(x)$ changes when we increase $x$ by $\Delta x$.

Try doing problem no. 2 yourself. To get started, let's take the case when $\Delta y=0$. For this to be true, we need $e^{\beta \Delta x}=1$, which only happens when $\Delta x=0$ (if $\beta\not=0$).