# How do you find the exponential model y=ae^(bx) that goes through the points (4,256) (3,64)?

##### 1 Answer
Jul 4, 2016

$y = {e}^{2 \ln \left(2\right) x} \cong {e}^{1.386 x}$

#### Explanation:

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:

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

$b = \ln \frac{256 / 64}{4 - 3} = \ln \frac{4}{1} = \ln \left({2}^{2}\right) = 2 \ln \left(2\right)$

plugging this into our initial equation we get

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

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

$64 = a \cdot {2}^{2 \cdot 3} = 64 a \implies a = 1$

So our equation becomes:

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

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

$y = {e}^{2 \ln \left(2\right) x} \cong {e}^{1.386 x}$