Question

How do you find the exponential model `y=ae^(bx)` that goes through the points (4,256) (3,64)?

Answer 1

`y=e^(2ln(2)x)~=e^(1.386x)`

Explanation:

One way to approach this problem is to do a substitution which changes the form of the equation to a line. We can do this by taking the `log` of both sides:

`ln(y) = bx +ln(a)`

In this equation, `b` is our slope which is given by:

`b=(ln(y_2)-ln(y_1))/(x_2-x_1) = ln(y_2//y_1)/(x_2-x_1)`

`b=ln(256//64)/(4-3)=ln(4)/1=ln(2^2)=2ln(2)`

plugging this into our initial equation we get

`y=ae^(2ln(2)x)=a[e^ln(2)]^(2x)=a2^(2x)`

Then we can get `a` by plugging our first point:

`64=a*2^(2*3)=64a implies a=1`

So our equation becomes:

`y=2^(2x)`

or, if we would like to maintain the exponential we would write:

`y=e^(2ln(2)x)~=e^(1.386x)`