# How do you find the exponential model y=ae^(bx) that goes through the points (5,3200) and (11,6800)?

Sep 2, 2016

$y = 3200 {\left(\frac{17}{8}\right)}^{\frac{x - 5}{6}}$.

#### Explanation:

Let us denote, by $C$, the curve  ; y=ae^(bx).

$C$ passes thro. the pts. $\left(5 , 3200\right) \mathmr{and} \left(11 , 6800\right)$, hence, the co-ords. of these pts. must satisfy the eq. of $C$.

$\therefore 3200 = a {e}^{5 b} \ldots \ldots \left(1\right) , \mathmr{and} , 6800 = a {e}^{11 b} \ldots \ldots \left(2\right)$

$\left(2\right) \div \left(1\right) \text{ gives, } \frac{6800}{3200} = \frac{a {e}^{11 b}}{a {e}^{5 b}}$

$\Rightarrow \frac{17}{8} = {e}^{6 b} \Rightarrow {\left(\frac{17}{8}\right)}^{\frac{1}{6}} = {e}^{b}$

Then, by $\left(1\right) , a = \frac{3200}{{e}^{b}} ^ 5 = \frac{3200}{\frac{17}{8}} ^ \left(\frac{5}{6}\right) = {\left(\frac{8}{17}\right)}^{\frac{5}{6}} 3200$

Hence, the reqd. expo. model is $: y = {\left(\frac{8}{17}\right)}^{\frac{5}{6}} 3200 {\left(\frac{17}{8}\right)}^{\frac{x}{6}}$,

=3200(17/8)^(x/6)*(17/8)^(-5/6, i.e.,

$y = 3200 {\left(\frac{17}{8}\right)}^{\frac{x - 5}{6}}$.

Enjoy maths.!