# Thread: Finding equation from logarithmic graph

1. ## Finding equation from logarithmic graph

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?

2. ## Re: Finding equation from logarithmic graph

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, $y= e^{4500}e^{-\frac{2000}{3}(x- 4)}$.