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

1 Answer
Jun 27, 2016

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)#