College Algebra, Chapter 5, 5.5, Section 5.5, Problem 6

The observed relative growth rate of Ukraine is $4\%$ per year. In 2006 the population of Ukraine is 112,000.

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

b.) Determine the projected population in year 2012.

c.) On what year will Ukraine have a total population of 200,000.




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

If we let $112000$ be the initial population, then the model of the population after $t$ years is $n(t) = 112000 e^{0.04t}$

b.) The projected population in the year 2012 where $t = 6$ is..


$
\begin{equation}
\begin{aligned}

n(6) =& 112000 e^{0.04(6)}
\\
\\
=& 142379.90 \text{ or } 142379

\end{aligned}
\end{equation}
$


c.)


$
\begin{equation}
\begin{aligned}

\text{if } n(t) =& 200,000 \text{ then}
&&
\\
\\
200,000 =& 112,000 e^{0.04 t}
&& \text{Divide both sides by } 112,000
\\
\\
\frac{25}{14} =& e^{0.04 t}
&& \text{Take $\ln$ of each side}
\\
\\
\ln \left( \frac{25}{14} \right) =& 0.04 t
&& \text{Recall } \ln e = 1
\\
\\
t =& \frac{\displaystyle \ln \left( \frac{25}{14} \right)}{0.04 }
&& \text{Solve for } t
\\
\\
t =& 14.50
&&
\end{aligned}
\end{equation}
$


It shows that the population will reach $200,000$ in the year 2020 plus 6 months.