It is nice that we are given the point, #(0,8),# because it allows us to find the value of #a# before we find the value of b:
Substitute the point #(0,8)# into #y=ae^(bx)#:
#8=ae^(b(0))#
Any number raised to the zero power is 1:
#8 = a(1)#
#a = 8#
Use the point, #(1,3),# to find the value of b:
#3 = 8e^(b(1))#
#e^b= 3/8#
#b = ln(3/8)#
The final equation is:
#y = 8e^(ln(3/8)x)#
Often, the same problem is asked where the x coordinate of one of the points is not 0. When this happens, you must find the value of #b# before you find the value of #a#; here is how you do it:
Given, two points, #(x_1,y_1)# and #(x_2,y_2)# and #y= ae^(bx)#
Write two equations by substituting each point into the given equation:
#y_1=ae^(bx_1)" [1]"#
#y_2=ae^(bx_2)" [2]"#
Divide equation [2] by equation [1]:
#y_2/y_1=(ae^(bx_2))/(ae^(bx_1))#
Please observe that #a# is eliminated because it is canceled by division:
#y_2/y_1=(cancel(a)e^(bx_2))/(cancel(a)e^(bx_1)) = (e^(bx_2))/(e^(bx_1))#
When you divide two numbers with the same base, it is the same as subtracting the exponents:
#y_2/y_1 = e^(bx_2-bx_1)#
Flip the equation and use the natural logarithm on both sides:
#ln(e^(bx_2-bx_1))= ln(y_2/y_1)#
Because #ln# and #e# are inverses only the exponent remains on the left:
#bx_2-bx_1= ln(y_2/y_1)#
Factor out #b#:
#b(x_2-x_1)= ln(y_2/y_1)#
Divide both sides by #(x_2-x_1)#:
#b= ln(y_2/y_1)/(x_2-x_1)#
Now that you have the value of #b#, you can substitute its value into either equation [1] or equation [2] to solve for the value of #a#.
