How do you find the exponential model $y = a e^{b x}$ $y=ae^(bx)$ that goes through the points (-3,2)(1,164)?

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Answer 1 · 2017-02-16T15:55:19

Exponential model is $y = 54.5 e^{1.10168 x}$ $y=54.5e^(1.10168x)$

As $y = a e^{b x}$ $y=ae^(bx)$ goes through $(- 3, 2)$ $(-3,2)$ ,

we have $2 = a e^{- 3 b}$ $2=ae^(-3b)$ ......................(1)

and as $y = a e^{b x}$ $y=ae^(bx)$ also goes through $(1, 164)$ $(1,164)$ ,

we have $164 = a e^{b}$ $164=ae^b$ ......................(2)

Dividing (2) by (1), we get

$\frac{e^{b}}{e^{- 3 b}} = \frac{164}{2} = 82$ $e^b/e^(-3b)=164/2=82$ or $e^{4 b} = 82$ $e^(4b)=82$ and $4 b = ln 82 = 4.40672$ $4b=ln82=4.40672$

and $b = \frac{4.40672}{4} = 1.10168$ $b=4.40672/4=1.10168$

Putting this value of $b$ $b$ in (2), we get

$a e^{1.10188} = 164$ $ae^1.10188=164$ or $3.0092 a = 164$ $3.0092a=164$

and $a = \frac{164}{3.0092} = 54.5$ $a=164/3.0092=54.5$

and hence exponential model is $y = 54.5 e^{1.10168 x}$ $y=54.5e^(1.10168x)$

How do you find the exponential model y=aebxy=ae^(bx) that goes through the points (-3,2)(1,164)?