Finding equation from logarithmic graph

Dec 2017
1
0
Kansas City
I've spent some time researching and trying to find an equation for this line, but it's not exact. I'm only searching for the equation of the line that descends towards zero (the angled line). I plugged in some numbers and it does not match the graph, the line on the graph is steeper. I start with y=a*b^x, solve for a and b then solve to show what x equals (commission). I followed some log rules and end up with the equation you see below in the image.

Can anyone help me find the equation?

Screen Shot 2017-12-22 at 4.07.13 PM.png
Screen Shot 2017-12-22 at 4.12.17 PM.png
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
The equation of a straight line that passes through $(x_0, y_0)$ and $(x_1, y_1)$ is $y= \frac{y_1- y_0}{x_1- x_0}(x- x_0)+ y_0$. The equation of a straight line that passes through (4, 4500) and (10, 500) is $y= \frac{500- 4500}{10- 4}(x- 4)+ 4500= -\frac{2000}{3}(x- 4)+ 4500$.

Since, here, the vertical axis gives log y rather than y, the equation is $log(y)= \frac{500- 4500}{10- 4}(x- 4)+ 4500= -\frac{2000}{3}(x- 4)+ 4500$. Taking the exponential of both sides, $y= e^{4500}e^{-\frac{2000}{3}(x- 4)}$.