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