Question

A town has a population of $12000$ and grows at $4\%$ every year. To the nearest year, how long will it be until the population will reach $26900$?

Answer 1

Step-1: Find population function:

Let the time after which the population of the town is $26900$ be $n$.

The current population of the town is $12000$ and it increases by $4\%$ every year.

The population of town after $n$ years is given by formula of amount of compound interest,

$A=P{\left(1+\frac{r}{100}\right)}^n$

Where, $A$ is population after $n$ years, $P$ is the current population of the city and $r$ is the rate of increase in population.

Here, $A=26900$, $P=12000$ and $r=4$.

Step-2: Find the time:

Substitute, $A=26900$, $P=12000$ and $r=4$ in $A=P{\left(1+\frac{r}{100}\right)}^n$.

$\begin{array}{rcl}26900& =& 12000{\left(1+\frac{4}{100}\right)}^n\\ & \Rightarrow & \frac{26900}{12000}={\left(1+0.04\right)}^n\left(\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}12000\right)\\ & \Rightarrow & 2.24={\left(1.04\right)}^n\\ & \Rightarrow & \ln \left(2.24\right)=\ln \left[{\left(1.04\right)}^n\right]\left(\mathrm{Take}\ln \mathrm{on}\mathrm{both}\mathrm{sides}\right)\\ & \Rightarrow & \ln \left(2.24\right)=n\left[\ln \left(1.04\right)\right]\left(\mathrm{Use}\ln a^n=n\ln a\right)\\ & \Rightarrow & 0.806=n·\left(0.039\right)\\ & \Rightarrow & \frac{0.806}{0.039}=n\left(\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}0.039\right)\\ & \Rightarrow & 20.67=n\\ \therefore n& \approx & 21\mathrm{years}\end{array}$

Hence, time required to reach the population by $26900$ is $21\mathrm{years}$.