Question

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

Answer 1

`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`.