Chapter 3, Problem 75RE

Solution

(a)

To calculate: Function that models the population as the function of time $t$ .

The function is $f(t)=89,000\times {0.982}^t$

Given information: Population $89000$ which is getting decreased by $1.8\%$ each year

Calculation: If the population decreases by $1.8\%$

each year, then each year there will be:

  $100\%-1.8\%=92.2\%$

  $92.2\%$ as many people as there were the year before.

Thus, each year the population is multiplied by $0.982$, according to this model:

  $\begin{array}{c}f\left(t\right)=89000\times 0.982\times \mathrm{0.982...}\times 0.982\\ =89000\times {0.982}^t\end{array}$

(b)

To calculate: Predicting when the population will be $50000$ .

Approx. $32\text{ years}$

Given information: Population $89000$ which is getting decreased by $1.8\%$ each year

Calculation:

As we know that $f\left(t\right)=89000\times {0.982}^t$ .

That returns population after $t$ years. We wish to find $t$ such that $f\left(t\right)=50000$. We substitute that in equation:

  $50000=89000\times {0.982}^t$

Isolate exponential:

  $\ln {0.982}^t=\ln \frac{50}{89}$

Take logarithm on both side and apply logarithmic properties:

  $\begin{array}{c}\ln {0.982}^t=\ln \left(\frac{50}{89}\right)\\ t\ln 0.982=\ln 50-\ln 89\\ t=\frac{\ln 50-\ln 89}{\ln 0.982}\\ t=31.75\text{ years}\end{array}$

Thus population should reach $50$ thousand in a little under $32\text{ years}$ .