Question

How do you find the exponential model `y=ae^(bx)` that goes through the points (2,8) (6, 128)?

Answer 1

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(128//8)/(6-2)=ln(16)/4=ln(2^4)/4=ln(2)`

plugging this into our initial equation we get

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

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

`8=a*2^2=4a implies a=2`

So our equation becomes:

`y=2*2^x`

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

`y=2e^(ln(2)x)~=2e^(0.6931x)`