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

Sep 20, 2016

$y = \frac{1}{2} {\left(\frac{1}{3}\right)}^{x} = \frac{1}{2 \cdot {3}^{x}} = 0.5 {\left(3\right)}^{-} x$.

#### Explanation:

The curve $C : y = a {e}^{b x} \text{ passes thro. the pts. } A \left(- 5 , \frac{243}{2}\right) , \mathmr{and} , B \left(- 1 , \frac{3}{2}\right) .$

$A \in C \Rightarrow \frac{243}{2} = a {e}^{- 5 b} \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)$.

$B \in C \Rightarrow \frac{3}{2} = a {e}^{- b} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$.

$\left(2\right) \div \left(1\right) \Rightarrow \frac{3}{2} \cdot \frac{2}{243} = {e}^{- b} \cdot {e}^{+ 5 b} , i . e . , {e}^{4 b} = \frac{1}{81}$

$\therefore {e}^{b} = {\left({3}^{-} 4\right)}^{\frac{1}{4}} = {3}^{-} 1 = \frac{1}{3}$.

$\text{By (2), then, } a = \frac{3}{2} \cdot {e}^{b} = \frac{3}{2} \cdot \frac{1}{3} = \frac{1}{2}$.

$\text{Therefore, C : } y = a {\left({e}^{b}\right)}^{x} = \frac{1}{2} {\left(\frac{1}{3}\right)}^{x} = \frac{1}{2 \cdot {3}^{x}} = 0.5 {\left(3\right)}^{-} x$.

Enjoy Maths.!