# The initial population of a large city is 20 million. Its population increases by 2% each year. It's 2012 today. How many people will live in the city at the end of 2016?

##### 1 Answer
Nov 27, 2016

a) An exponential model will be of the form $p = a \times {r}^{n}$, where the population after $n$ years is $p$ and the original population is $a$. The rate at which it increases (in decimals) is $r$.

Our equation, therefore is $p = 20 , 000 , 000 {\left(1.02\right)}^{n}$

b) To find the population after $2016$, we need to insert $n = 4$, because the time between the end of $2012$ and the end of $2016$ is $4$ years.

$p = 20 , 000 , 000 {\left(1.02\right)}^{4}$

$p = 21 , 648 , 643$

Hence, the population at the end of $2016$ will be $21 , 648 , 643$.

c) Set $p$ to $25 , 000 , 000$ and solve.

$p = 20 , 000 , 000 {\left(1.02\right)}^{n}$

$25 , 000 , 000 = 20 , 000 , 000 {\left(1.02\right)}^{n}$

$1.25 = {\left(1.02\right)}^{n}$

$\ln \left(1.25\right) = n \left(\ln 1.02\right)$

$n = \ln \frac{1.25}{\ln} \left(1.02\right)$

$n \cong 11.268$

However, we can't round to $11$ since after $11$ years, we won't have attained $25 , 000 , 000$ yet. After 12 years we will have, though. So, in the year $2012 + 12$, or $2024$, Nigeria will obtain a population of $25 , 000 , 000$

d) The doubling population will be $40 , 000 , 000$ (since 20,000,000 xx 2 = 40,000,000)

$40 , 000 , 000 = 20 , 000 , 000 {\left(1.02\right)}^{n}$

$2 = {\left(1.02\right)}^{n}$

$\ln 2 = n \ln \left(1.02\right)$

$n = \ln \frac{2}{\ln} \left(1.02\right)$

$n = 35.003$

Again, as with part c), we're going to have to use $n = 36$ since when $n = 35$, the population will not have yet doubled.

After $36$ years, we will be in $2012 + 36 = 2048$.

Hopefully this helps!