# How do you find the exponential model y=ae^(bx) that goes through the points (-1,125) (2,5)?

Jun 29, 2016

$y = {5}^{\frac{7}{3}} {e}^{- \frac{2 \ln 5}{3}} = {5}^{\frac{7 - 2 x}{3}}$

#### Explanation:

As $y = a {e}^{b x}$ passes through $\left(- 1 , 125\right)$ and $\left(2 , 5\right)$, hence

$125 = a {e}^{- b}$ i.e. $a = 125 {e}^{b}$ ................................(A)

and $5 = a {e}^{2 b}$ i.e. $a = \frac{5}{e} ^ \left(2 b\right)$ ................................(B)

Hence $125 {e}^{b} = \frac{5}{e} ^ \left(2 b\right)$ i.e. ${e}^{3 b} = \frac{5}{125} = \frac{1}{25}$

Hence $3 b = - \ln 25$ and $b = - \ln \frac{25}{3} = - 2 \ln \frac{5}{3}$

As from (B), ${e}^{2 b} = \frac{5}{a}$, putting this in (A),

$a = 125 \times {\left(\frac{5}{a}\right)}^{\frac{1}{2}}$ or ${a}^{\frac{3}{2}} = {5}^{\frac{7}{2}}$ and $a = {5}^{\frac{7}{3}}$

Hence equation is $y = {5}^{\frac{7}{3}} {e}^{- x \left(\frac{2 \ln 5}{3}\right)}$

Note that this can be simplified as $y = {5}^{\frac{7}{3}} {\left({e}^{\ln} 5\right)}^{- \left(\frac{2}{3} x\right)} = {5}^{\frac{7}{3}} \times {5}^{- \frac{2}{3} x} = {5}^{\frac{7 - 2 x}{3}}$