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

1 Answer
Ratnaker Mehta
Sep 20, 2016

y=1/2(1/3)^x=1/(2*3^x)=0.5(3)^-xy=12(13)x=12⋅3x=0.5(3)−x.

Explanation:

The curve C : y=ae^(bx)" passes thro. the pts. "A(-5,243/2), and, B(-1,3/2).C:y=aebx passes thro. the pts. A(−5,2432),and,B(−1,32).

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.!

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