What is the equation using ae^rx model and the points (-3,800) and (1,50)?

1 Answer
May 15, 2018

f(x) = 100 e^(-ln(2)x)

Explanation:

So we have the function f(x) = a e^(rx).

Let's plug in the two numbers that we know:
f(-3) = 800 = a e^(-3r)
f(1)\ \ \ \ \ = 50\ \ = a e^(r)
We have two equations and two unknowns, so we can solve for a, r. Let's divide the first equation by the second:
f(-3)/(f(1)) = 800/50 = 16 = (ae^(-3r))/(ae^r) = e^(-4r)
i.e. 16 = e^(-4r) implies -4r = ln(16) = ln(2^4) = 4ln2
So we have r = - ln(2) .

Plugging that into the second equation (it could also be the first, the second just seems easier):
50 = a e^(-ln2) = a(e^(ln(2)))^-1 = a*2^-1 implies a = 100
If we plug this into the first equation, we see that this also works:
f(-3) = 100 * e^(-3cdot - ln(2)) = 100e^(ln(2^3)) = 800

In summary, we plug in each of the values and solve to find a = 100, r = -ln2.