How do you find the exponential model #y=ae^(bx)# that goes through the points (0,7) (5, 32)?

2 Answers
Aug 31, 2016

#y = 7e^(1/5ln(32/7)x)#

Explanation:

We can solve for #a# without writing a systems of equations. Notice in the first point, #x = 0#. This will make #b = 0# since #0 xx a = 0#. We will be left with only #a#.

#y = ae^(bx)#

#7 = ae^(b xx 0)#

#7 = ae^0#

#7 = a(1)#

#a = 7#

Substituting, and solving for b:

#y = ae^(bx)#

#32 = 7e^(5b)#

#32/7 = e^(5b)#

#ln(32/7) = ln(e^(5b))#

#ln(32/7) = 5b(lne)#

#ln(32/7) = 5b#

#(ln(32/7))/5 = b#

#1/5ln(32/7) = b#

Hence, the equation of the function is

#y = 7e^(1/5ln(32/7)x)#

Hopefully this helps!

Aug 31, 2016

#y=7(2*7^(-1/5))^x#.

Explanation:

Let us denote, by #C#, the curve represented by # : y=ae^(bx)...(0)#.

The pt. #(0,7) in C rArr "its co-ords. must satisfy (0)"#.

#rArr 7=ae^0=a.........(1)#.

#" pt."(5,32) in C rArr 32=ae^(5b)#.

By #(1) "then", e^(5b)=32/7, or, e^b=(32/7)^(1/5)#.

Hence, #(0)# becomes,

#y=ae^(bx)=7(e^b)^x=7{(32/7)^(1/5)}^x=7(2*7^(-1/5))^x#.