# How do you find the exponential model y=ae^(bx) that goes through the points (2,8) (6, 128)?

Jun 27, 2016

One way to approach this problem is to do a substitution which changes the form of the equation to a line. We can do this by taking the $\log$ of both sides:

$\ln \left(y\right) = b x + \ln \left(a\right)$

In this equation, $b$ is our slope which is given by:

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

$b = \ln \frac{128 / 8}{6 - 2} = \ln \frac{16}{4} = \ln \frac{{2}^{4}}{4} = \ln \left(2\right)$

plugging this into our initial equation we get

$y = a {e}^{\ln \left(2\right) x} = a {\left[{e}^{\ln} \left(2\right)\right]}^{x} = a {2}^{x}$

Then we can get $a$ by plugging our first point:

$8 = a \cdot {2}^{2} = 4 a \implies a = 2$

So our equation becomes:

$y = 2 \cdot {2}^{x}$

or, if we would like to maintain the exponential we would write:

$y = 2 {e}^{\ln \left(2\right) x} \cong 2 {e}^{0.6931 x}$