# How do you find the exponential model y=ae^(bx) that goes through the points (1, 1.2) and (4, 0.0768)?

Jul 19, 2016

$y = 3 {\left(0.4\right)}^{x}$

#### Explanation:

The Exponential curve $y = a {e}^{b x}$ passes thro. pts. $\left(1 , 1.2\right)$ and

$\left(4 , 0.0768\right)$

Accordingly, the respective co-ords. must satisfy the eqn. of the curve.

$\therefore 1.2 = a {e}^{b} , \mathmr{and} , 0.0768 = a {e}^{4 b} \Rightarrow \frac{0.0768}{1.2} = \frac{a {e}^{4 b}}{a {e}^{b}}$

$\Rightarrow 0.064 = {e}^{3 b} = {\left({e}^{b}\right)}^{3} \Rightarrow {0.064}^{\frac{1}{3}} = {\left\{{\left({e}^{b}\right)}^{3}\right\}}^{\frac{1}{3}}$

$\Rightarrow 0.4 = {e}^{b}$. Then, from $1.2 = a {e}^{b}$, we have, $1.2 = 0.4 a$

$\Rightarrow a = 3$

Hence, the reqd. eqn.$: y = a {e}^{b x} = a {\left({e}^{b}\right)}^{x}$,i.e., $y = 3 {\left(0.4\right)}^{x}$