Question #5bf51

1 Answer
Jim G.
Jan 8, 2018

#y=1.418e^(0.043x)#

Explanation:

#"substitute "(0,1.418)" into "y=ae^(bx)#

#rArr1.418=ae^0rArra=1.418#

#"substitute "(8,2.001)" into "y=ae^(bx)#

#rArr2.001=1.418e^(8b)#

#rArre^(8b)=2.001/1.418#

#[logx^nhArrnlogx]#

#rArr8bcancel(lne)=ln(2.001/1.418)#

#rArrb=(ln(2.001/1.418))/8~~0.043" to 3 dec. places"#

#rArry=1.418e^(0.043x)#

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