College Algebra, Chapter 5, 5.5, Section 5.5, Problem 12

In 1950, the population California was 10,586,223 and in 1980, the population is 23,668,562. Assuming that the population of California grows exponentially.

a.) Determine a function that will model the populations t years after 1950.

b.) Identify the needed time for the population to double.

c.) By using the function from part(a), predict the population of Florida in the year 2006.



a.) Recall the formula for growth rate

$n(t) = no e^{rt}$

where

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

$n_0$ = initial size of the population

$r$ = relative rate of growth

$t$ = time

If we let the population in 1950 be the initial population. And in 1980, ($t = 30$ years from 1950) the population is 23668562. So..


$
\begin{equation}
\begin{aligned}

23,668,562 =& 10, 586,223 e^{r(30)}
&& \text{Divide both sides by } 10,586, 223
\\
\\
2.2358 =& e^{30 r}
&& \text{Take $\ln$ of each side}
\\
\\
\ln (2.2358) =& 30r
&& \text{Recall that } \ln e =1
\\
\\
r =& \frac{\ln (2.2358)}{30}
&& \text{Divide each side by } 30
\\
\\
r =& 0.0268
&& \text{Solve for } x
\\
\\
r =& 2.68 \%
&& \text{Express as percentage}

\end{aligned}
\end{equation}
$


Thus, the model is represented as

$n(t) = 10, 586,223 e^{0.0268 t}$

b.) If the population is to double then $n = 2n_0$, so


$
\begin{equation}
\begin{aligned}

2n_0 =& n_0 e^{0.0268 t}
&& \text{Divide each side by } n_0
\\
\\
2 =& e^{0.0268 t}
&& \text{Take $\ln$ of both sides}
\\
\\
\ln 2 =& 0.0268 t
&& \text{Recall that } \ln e = 1
\\
\\
t =& \frac{\ln 2}{0.0268}
&& \text{Solve for } t
\\
\\
t =& 25.8445
&&

\end{aligned}
\end{equation}
$


It shows that the population will double after 26 years from 1950 or in the year 1976.

c.) In the year 2006, at $t = 56$, the population will be


$
\begin{equation}
\begin{aligned}

n(56) =& 10, 586, 223 e^{0.0268(56)}
\\
\\
n(56) =& 47,482,130.43 \text{ or } 47,482,130

\end{aligned}
\end{equation}
$