# How do you find the exponential model y=ae^(bx) that goes through the points (0,7) (5, 32)?

##### 2 Answers
Aug 31, 2016

$y = 7 {e}^{\frac{1}{5} \ln \left(\frac{32}{7}\right) x}$

#### Explanation:

We can solve for $a$ without writing a systems of equations. Notice in the first point, $x = 0$. This will make $b = 0$ since $0 \times a = 0$. We will be left with only $a$.

$y = a {e}^{b x}$

$7 = a {e}^{b \times 0}$

$7 = a {e}^{0}$

$7 = a \left(1\right)$

$a = 7$

Substituting, and solving for b:

$y = a {e}^{b x}$

$32 = 7 {e}^{5 b}$

$\frac{32}{7} = {e}^{5 b}$

$\ln \left(\frac{32}{7}\right) = \ln \left({e}^{5 b}\right)$

$\ln \left(\frac{32}{7}\right) = 5 b \left(\ln e\right)$

$\ln \left(\frac{32}{7}\right) = 5 b$

$\frac{\ln \left(\frac{32}{7}\right)}{5} = b$

$\frac{1}{5} \ln \left(\frac{32}{7}\right) = b$

Hence, the equation of the function is

$y = 7 {e}^{\frac{1}{5} \ln \left(\frac{32}{7}\right) x}$

Hopefully this helps!

Aug 31, 2016

$y = 7 {\left(2 \cdot {7}^{- \frac{1}{5}}\right)}^{x}$.

#### Explanation:

Let us denote, by $C$, the curve represented by $: y = a {e}^{b x} \ldots \left(0\right)$.

The pt. $\left(0 , 7\right) \in C \Rightarrow \text{its co-ords. must satisfy (0)}$.

$\Rightarrow 7 = a {e}^{0} = a \ldots \ldots \ldots \left(1\right)$.

$\text{ pt.} \left(5 , 32\right) \in C \Rightarrow 32 = a {e}^{5 b}$.

By $\left(1\right) \text{then} , {e}^{5 b} = \frac{32}{7} , \mathmr{and} , {e}^{b} = {\left(\frac{32}{7}\right)}^{\frac{1}{5}}$.

Hence, $\left(0\right)$ becomes,

$y = a {e}^{b x} = 7 {\left({e}^{b}\right)}^{x} = 7 {\left\{{\left(\frac{32}{7}\right)}^{\frac{1}{5}}\right\}}^{x} = 7 {\left(2 \cdot {7}^{- \frac{1}{5}}\right)}^{x}$.