College Algebra, Chapter 5, 5.5, Section 5.5, Problem 8

Initially there are 1500 bacteria in a culture and it double every 30
minutes.

a.) Determine a function that will model the number of bacteria n(t) after t minutes.

b.) Determine the number of bacteria after 2 hours

c.) After how many minutes will the culture contain 4000 bacteria?



a.) Recall the formula for growth rate

$n(t) = n_0 e^{rt}$

where

$n(t)$ = population at time $t$

$n_0$ = initial size of the population

$r$ = relative rate of growth

$t$ = time

If the population doubles every 30 min (0.5 hours) then $n = 2n_0$


$
\begin{equation}
\begin{aligned}

2n_0 =& n_0 e^{r(0.5)}
&& \text{Divide both sides by } n_0
\\
\\
2 =& e^{r(0.5)}
&& \text{Take $\ln$ of each side}
\\
\\
\ln 2 =& r(0.5)
&& \text{Recall that } \ln e = 1
\\
\\
r =& \frac{\ln 2}{0.5}
&& \text{Solve for } r
\\
\\
r =& 1.3863
&&
\end{aligned}
\end{equation}
$


Therefore, the model is represented as

$n(t) = 1500 e ^{1.3863 t}$

b.)


$
\begin{equation}
\begin{aligned}

\text{if } t =& 2 \text{ hours, then}
\\
\\
n(2) =& 1500 e^{1.3863(2)}
\\
\\
=& 2400

\end{aligned}
\end{equation}
$


c.)


$
\begin{equation}
\begin{aligned}

\text{if } n(t) =& 4000 \text{ then}
&&
\\
\\
4000 =& 1500 e^{1.3863 (t)}
&& \text{Divide both sides by } 1500
\\
\\
\frac{8}{3} =& e^{1.3863 t}
&& \text{Take $\ln$ of each side}
\\
\\
\ln \left( \frac{8}{3} \right) =& 1.3863 t
&& \text{Recall that } \ln e = 1
\\
\\
t =& \frac{\displaystyle \ln \left( \frac{8}{3} \right) }{1.3863}
&& \text{Solve for } t
\\
\\
t =& 0.7075 \text{ hours } \times \frac{60 \text{ minutes}}{1 \text{ hour}}
&& \text{Convert hours into minutes}
\\
\\
t =& 42.45 \text{ minutes }
&&
\end{aligned}
\end{equation}
$


It shows that the population of bacteria will be 4000 after 43 minutes.