Question

Population Decline

The population of a midwestern city follows the exponential law.

(a) If N is the population of the city and t is the time in years, express N as a function of t.

(b) If the population decreased from 900,000 to 800,000 from 2005 to 2007, what will the population be in 2009?

Short answer

(a) N as a function of t can be expressed as $N\left(t\right)=N_0{e}^{kt}$

(b) Population in 2009 will be 697007

Step-by-step solution

Step 1. Given information

The population of a midwestern city follows the exponential law.

Part (a) of Step 1. N as a function of t 

If N is the population of the city and t is the time in years, then N as a function of t can be expressed as $N\left(t\right)=N_0{e}^{kt}$, where $N_0$is the initial population, $N\left(t\right)$is the current population, t is time and $k$is positive constant.

Part (b) of Step 1. Population is 2009

The population decreased from 900,000 to 800,000 from 2005 to 2007

$$\begin{gathered} N_0=900,000, \\ N\left(t\right)=800,000, \\ t=2 \end{gathered}$$

Substitute the values of $N_0,N\left(t\right),t$in the function role="math" localid="1647288977383" $N\left(t\right)=N_0{e}^{kt}$

$$\begin{gathered} N\left(t\right)=N_0{e}^{kt} \\ 800,000=900,000e^{k\times 2} \\ \frac{8}{9}=e^{k\times 2} \\ \ln \left(\frac{8}{9}\right)=\ln \left(e^{k\times 2}\right) \\ \ln \left(\frac{8}{9}\right)=k\times 2 \\ k=\frac{\ln \left(\frac{8}{9}\right)}{2} \\ k=-0.0639 \end{gathered}$$

Now population in 2009 can be calculated by

substituting $N_0=900,000,k=-0.0639,t=4$in the function $N\left(t\right)=N_0{e}^{kt}$

$$\begin{gathered} N\left(t\right)=N_0{e}^{kt} \\ N\left(2009\right)=900,000e^{-0.0639\times 4} \\ N\left(2009\right)=697,007 \end{gathered}$$