How do you find the exponential model #y=ae^(bx)# that goes through the points (1, 1.2) and (4, 0.0768)?

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Answer 1 · 2016-07-19T19:05:56

#y=3(0.4)^x#

The Exponential curve #y=ae^(bx)# passes thro. pts. #(1,1.2)# and

#(4,0.0768)#

Accordingly, the respective co-ords. must satisfy the eqn. of the curve.

#:.1.2=ae^b, and, 0.0768=ae^(4b) rArr 0.0768/1.2=(ae^(4b))/(ae^b)#

#rArr 0.064=e^(3b)=(e^b)^3 rArr 0.064^(1/3)={(e^b)^3}^(1/3)#

#rArr 0.4=e^b#. Then, from #1.2=ae^b#, we have, #1.2=0.4a#

#rArr a=3#

Hence, the reqd. eqn.# : y=ae^(bx)=a(e^b)^x#,i.e., #y=3(0.4)^x#