College Algebra, Chapter 5, 5.5, Section 5.5, Problem 10

There are 400 bacteria's after 2 hours and 25,600 after 6 hours in a culture.

a.) Identify the relative growth rate of the bacteria population? Express your answer in percentage.

b.) Determine the initial size of the culture?

c.) Find a function that will model the number of bacteria n(t) after t hours.

d.) What will be the number of bacteria after 4.5hours?

e.) When will the number of bacteria be 50,000?



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



$
\begin{equation}
\begin{aligned}

\text{if } n(2) =& 400, \text{ then}
&& \text{and}&
\text{if } n(6) =& 25600 \text{ then}
\\
\\
400 =& n_0 e^{r(2)}
&& &
25600 =& n_0 e^{r(6)}
\\
\\
n_0 =& \frac{400}{e^{2r}} \qquad \text{Equation 1}
&& &
n_0 =& \frac{25600}{e^{6r}} \qquad \text{Equation 2}

\end{aligned}
\end{equation}
$


By using equations 1 and 2,


$
\begin{equation}
\begin{aligned}

\frac{400}{e^{2r}} =& \frac{25600}{e^{6r}}
&& \text{Multiply both sides by $e^{6r}$ and divide each side by } 400
\\
\\
\frac{e^{6r}}{e^{2r}} =& \frac{25600}{400}
&& \text{Apply Property of Exponents}
\\
\\
e^{4r} =& 64
&& \text{Take $\ln$ of each sides}
\\
\\
4r =& \ln (64)
&& \text{Recall } \ln e = 1
\\
\\
r =& \frac{\ln (64)}{4}
&& \text{Solve for } r
\\
\\
r =& 1.0397 \times 100 \%
&& \text{Express as a percentage}
\\
\\
r =& 103.97 \%
&&

\end{aligned}
\end{equation}
$


b.) By using equation 1


$
\begin{equation}
\begin{aligned}

n_0 =& \frac{400}{e^{2(1.0397)}}
\\
\\
n_0 =& 50

\end{aligned}
\end{equation}
$


c.) By substituting all the acquired information in the general equation, we have

$n(t) = 50e^{1.0397 t}$

d.)


$
\begin{equation}
\begin{aligned}

\text{if } t =& 4.5 \text{ hours, then}
\\
\\
n(4.5) =& 50 e^{1.0397 (4.5)}
\\
\\
n(4.5) =& 5381.23 \text{ or } 5381

\end{aligned}
\end{equation}
$


e.)


$
\begin{equation}
\begin{aligned}

\text{if } n(t) = 50,000 \text{ then}
&&
\\
\\
50,000 =& 50 e^{1.0397 (t)}
&& \text{Divide each side by } 50
\\
\\
1000 =& e^{1.0397(t)}
&& \text{Take $\ln$ of each side}
\\
\\
\ln (1000) =& 1.0397 t
&& \text{Recall } \ln e = 1
\\
\\
t =& \frac{\ln (1000)}{1.0397}
&& \text{Solve for } t
\\
\\
t =& 6.64 \text{ hours}

\end{aligned}
\end{equation}
$