Chapter 5.6, Problem 98AYU

A Population Model The population of the world in 2015 was $7.21$ billion people and was growing at a rate of $1.1\%$ per year. Assuming that this growth rate continues, the model $P\left(t\right)=7.21{\left(1.011\right)}^{t-2015}$ represents the population $P$ (in billions of people) in year $t$.

(a) According to this model, when will the population of the world be 9 billion people?

(b) According to this model, when will the population of the world be $12.5$ billion people?

Source: U.S Census Bureau

To determine

To find:

a. According to this model, when will the population of the world be 9 billion people?

Answer to Problem 98AYU

a. 2042

Explanation of Solution

Given:

The population of the world in 2015 was $7.21$ billion people and was growing at a rate of $1.1\%$ per year. Assuming that this growth rate continues, the model $P\left(t\right)=7.21{\left(1.011\right)}^{t-2015}$ represents the population $\text{P}$ (in billions of people) in year $t$.

Calculation:

$P\left(t\right)=7.21{\left(1.011\right)}^{t-2015}$

a. Determine when will the population of the world be 9 billion people?

$9=7.21{\left(1.011\right)}^{t-2015}$

$\frac{9}{7.21}={\left(1.011\right)}^{t-2015}$

$1.2483={\left(1.011\right)}^{t-2015}$

$\log 1.2483=\log {\left(1.011\right)}^{t-2015}$

$\log 1.2483=t-2015\log 1.011$

$\frac{\log 1.2483}{\log 1.011}=t-2015$

$20.27=t-2015$

$t=2042$

To determine

To find:

b. According to this model, when will the population of the world be $12.5$ billion people?

Answer to Problem 98AYU

b. 2065

Explanation of Solution

Given:

The population of the world in 2015 was $7.21$ billion people and was growing at a rate of $1.1\%$ per year. Assuming that this growth rate continues, the model $P\left(t\right)=7.21{\left(1.011\right)}^{t-2015}$ represents the population $\text{P}$ (in billions of people) in year $t$.

Calculation:

$P\left(t\right)=7.21{\left(1.011\right)}^{t-2015}$

b. Determine when will the population of the world be $12.5$ billion people?

$12.5=7.21{\left(1.011\right)}^{t-2015}$

$\frac{12.5}{7.21}={\left(1.011\right)}^{t-2015}$

$1.7337={\left(1.011\right)}^{t-2015}$

$1.7337=\log {\left(1.011\right)}^{t-2015}$

$\log 1.7337=t-2015\log 1.011$

$\frac{\log 1.7337}{\log 1.011}=t-2015$

$50.2982=t-2015$

$t=2065$