# A serving of a popular cola soft drink has 46 mg of caffeine. Suppose the half-life for caffeine remaining in he body of a typical adult is 6 h. How do you write an exponential decay function that gives the amount of caffeine in the body after t hours?

Nov 27, 2017

See below.

#### Explanation:

We need an equation of the form:

$y \left(t\right) = a {e}^{k} t$

Where $a$ is the initial amount, $k$ is the growth/decay constant and $t$ is the time, in this case hours.

To find the the constant $k$, we need to know the initial amount $a$

$a = 46 m g$

We know the half life is 6 hours, so after 6 hours the amount $y$ will be $23 m g$

So:

$y \left(t\right) = a {e}^{k} t$

$23 = 46 {e}^{6} k$

Solve for $k$:

$\frac{23}{46} = {e}^{6 k}$

Taking natural logs of both sides:

$\ln \left(\frac{23}{46}\right) = 6 k \ln e \textcolor{w h i t e}{88}$ ( $\ln e = 1$ )

$k = \frac{1}{6} \ln \left(\frac{23}{46}\right)$

So our equation is:

$y \left(t\right) = 46 {e}^{\frac{1}{6} \ln \left(\frac{23}{46}\right) t}$

This could also be written:

$y \left(t\right) = 46 {e}^{\frac{t}{6} \ln \left(\frac{23}{46}\right)}$

TEST:

$t = 6$

$y \left(t\right) = 46 {e}^{\frac{1}{6} \ln \left(\frac{23}{46}\right) 6} = 23$

Graph of $y \left(t\right) = 46 {e}^{\frac{1}{6} \ln \left(\frac{23}{46}\right) t}$ 