How do you find the exponential model #y=ae^(bx)# that goes through the points (-1,125) (2,5)?

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Answer 1 · 2016-06-29T04:57:31

#y=5^(7/3)e^(-(2ln5)/3)=5^((7-2x)/3)#

As #y=ae^(bx)# passes through #(-1,125)# and #(2,5)#, hence

#125=ae^(-b)# i.e. #a=125e^b# ................................(A)

and #5=ae^(2b)# i.e. #a=5/e^(2b)# ................................(B)

Hence #125e^b=5/e^(2b)# i.e. #e^(3b)=5/125=1/25#

Hence #3b=-ln25# and #b=-ln25/3=-2ln5/3#

As from (B), #e^(2b)=5/a#, putting this in (A),

#a=125xx(5/a)^(1/2)# or #a^(3/2)=5^(7/2)# and #a=5^(7/3)#

Hence equation is #y=5^(7/3)e^(-x((2ln5)/3))#

Note that this can be simplified as #y=5^(7/3)(e^ln5)^(-(2/3x))=5^(7/3)xx5^(-2/3x)=5^((7-2x)/3)#