How do you find the exponential model #y=ae^(bx)# that goes through the points (-5, 243/2) and (-1, 3/2)?

Question

1292 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-20T17:23:54

#y=1/2(1/3)^x=1/(2*3^x)=0.5(3)^-x#.

The curve # C : y=ae^(bx)" passes thro. the pts. "A(-5,243/2), and, B(-1,3/2).#

# A in C rArr 243/2=ae^(-5b)..................(1)#.

# B in C rArr 3/2=ae^(-b)......................(2)#.

# (2)-:(1) rArr 3/2*2/243=e^(-b)*e^(+5b), i.e., e^(4b)=1/81#

#:. e^b=(3^-4)^(1/4)=3^-1=1/3#.

#"By (2), then, "a=3/2*e^b=3/2*1/3=1/2#.

#"Therefore, C : "y=a(e^b)^x=1/2(1/3)^x=1/(2*3^x)=0.5(3)^-x#.

Enjoy Maths.!