# How do you find the exponential model y = ae^(bx) that fits the two points (0, 8), (1, 3)?

Mar 25, 2018

It is nice that we are given the point, $\left(0 , 8\right) ,$ because it allows us to find the value of $a$ before we find the value of b:

Substitute the point $\left(0 , 8\right)$ into $y = a {e}^{b x}$:

$8 = a {e}^{b \left(0\right)}$

Any number raised to the zero power is 1:

$8 = a \left(1\right)$

$a = 8$

Use the point, $\left(1 , 3\right) ,$ to find the value of b:

$3 = 8 {e}^{b \left(1\right)}$

${e}^{b} = \frac{3}{8}$

$b = \ln \left(\frac{3}{8}\right)$

The final equation is:

$y = 8 {e}^{\ln \left(\frac{3}{8}\right) x}$

Often, the same problem is asked where the x coordinate of one of the points is not 0. When this happens, you must find the value of $b$ before you find the value of $a$; here is how you do it:

Given, two points, $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ and $y = a {e}^{b x}$

Write two equations by substituting each point into the given equation:

${y}_{1} = a {e}^{b {x}_{1}} \text{ [1]}$

${y}_{2} = a {e}^{b {x}_{2}} \text{ [2]}$

Divide equation [2] by equation [1]:

${y}_{2} / {y}_{1} = \frac{a {e}^{b {x}_{2}}}{a {e}^{b {x}_{1}}}$

Please observe that $a$ is eliminated because it is canceled by division:

${y}_{2} / {y}_{1} = \frac{\cancel{a} {e}^{b {x}_{2}}}{\cancel{a} {e}^{b {x}_{1}}} = \frac{{e}^{b {x}_{2}}}{{e}^{b {x}_{1}}}$

When you divide two numbers with the same base, it is the same as subtracting the exponents:

${y}_{2} / {y}_{1} = {e}^{b {x}_{2} - b {x}_{1}}$

Flip the equation and use the natural logarithm on both sides:

$\ln \left({e}^{b {x}_{2} - b {x}_{1}}\right) = \ln \left({y}_{2} / {y}_{1}\right)$

Because $\ln$ and $e$ are inverses only the exponent remains on the left:

$b {x}_{2} - b {x}_{1} = \ln \left({y}_{2} / {y}_{1}\right)$

Factor out $b$:

$b \left({x}_{2} - {x}_{1}\right) = \ln \left({y}_{2} / {y}_{1}\right)$

Divide both sides by $\left({x}_{2} - {x}_{1}\right)$:

$b = \ln \frac{{y}_{2} / {y}_{1}}{{x}_{2} - {x}_{1}}$

Now that you have the value of $b$, you can substitute its value into either equation [1] or equation [2] to solve for the value of $a$.