Question

The population of the world in 2018 was 7.63 billion people and was growing at a rate of \(1.1 \%\) per year. Assuming that this growth rate continues, the model \(P(t)=7.63(1.011)^{t-2018}\) 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?

Short answer

The population will be 9 billion in approximately 2031, and 12.5 billion in approximately 2055.

Step-by-step solution

Understand the given model

The population model given is \[ P(t) = 7.63(1.011)^{t-2018} \] where \( P(t) \) represents the population in billions in the year \( t \), starting from 2018.

Set up the equation for part (a)

For part (a), we want to find the year \( t \) when the population is 9 billion. Set up the equation by substituting \( P(t) = 9 \): \[ 9 = 7.63(1.011)^{t-2018} \]

Solve the equation for part (a)

Divide both sides by 7.63: \[ \frac{9}{7.63} = (1.011)^{t-2018} \] Calculate the left-hand side: \[ 1.178 = (1.011)^{t-2018} \] Take the natural logarithm of both sides to solve for \( t \): \[ \text{ln}(1.178) = \text{ln}((1.011)^{t-2018}) \] Simplify using the properties of logarithms: \[ \text{ln}(1.178) = (t-2018) \text{ln}(1.011) \] Solve for \( t \): \[ t - 2018 = \frac{ \text{ln}(1.178) }{ \text{ln}(1.011) } \] Calculate the right-hand side and then add 2018 to isolate \( t \): \[ t = \frac{ \text{ln}(1.178) }{ \text{ln}(1.011) } + 2018 \] Using a calculator, calculate the value: \[ t ≈ 2031 \]

Set up the equation for part (b)

For part (b), we want to find the year \( t \) when the population is 12.5 billion. Set up the equation by substituting \( P(t) = 12.5 \): \[ 12.5 = 7.63(1.011)^{t-2018} \]

Solve the equation for part (b)

Divide both sides by 7.63: \[ \frac{12.5}{7.63} = (1.011)^{t-2018} \] Calculate the left-hand side: \[ 1.638 = (1.011)^{t-2018} \] Take the natural logarithm of both sides to solve for \( t \): \[ \text{ln}(1.638) = \text{ln}((1.011)^{t-2018}) \] Simplify using the properties of logarithms: \[ \text{ln}(1.638) = (t-2018) \text{ln}(1.011) \] Solve for \( t \): \[ t - 2018 = \frac{ \text{ln}(1.638) }{ \text{ln}(1.011) } \] Calculate the right-hand side and then add 2018 to isolate \( t \): \[ t = \frac{ \text{ln}(1.638) }{ \text{ln}(1.011) } + 2018 \] Using a calculator, calculate the value: \[ t ≈ 2055 \]

Key concepts

Population Modeling

Population modeling is a way to predict how populations grow over time. This can be for humans, animals, or even bacteria. Using models, scientists and researchers can make predictions about the future.
In the given exercise, we're modeling the world's population growth. The model given is $$ P(t) = 7.63(1.011)^{t-2018} $$ Here, $$ P(t) $$ represents the population in billions, and $$ t $$ is the year. This model helps us understand how the population changes over time, assuming a constant growth rate.
Using this model, we can estimate when the population will reach certain milestones, such as 9 billion or 12.5 billion people.

Logarithms

Logarithms are a mathematical tool used to solve equations involving exponents. They play a key role in population modeling.
For example, in our exercise, to find the year when the population will be a certain number, we need to solve for $$ t $$ in the equation:
$$ 9 = 7.63(1.011)^{t-2018} $$By taking the natural logarithm (ln) of both sides, we can simplify and solve for $$ t $$:

$$ \text{ln}(1.178) = (t-2018) \text{ln}(1.011) $$ This makes it easier to isolate $$ t $$. Logarithms are very handy when dealing with exponential functions, as they transform multiplicative processes into additive ones, making complex equations more manageable.

Growth Rate Calculations

Calculating growth rates is crucial for understanding how fast a population increases. In our exercise, the growth rate is given as $$ 1.1\things% $$ per year. This means each year, the population grows by 1.1% of its current size.
The growth rate is used in the model formula:
$$ P(t) = 7.63(1.011)^{t-2018} $$
Here, 1.011 represents a 1.1% growth rate. To understand how this impacts population growth, imagine starting with 7.63 billion people:

• After 1 year: $$ 7.63 \times 1.011 $$
• After 2 years: $$ 7.63 \times 1.011^2 $$
• And so on...

This exponential growth shows how a small growth rate can result in large population increases over time.

Exponential Functions

Exponential functions describe situations where a quantity grows at a rate proportional to its current value. In our exercise, the population growth is modeled by an exponential function:
$$ P(t) = 7.63(1.011)^{t-2018} $$ Here, $$ 1.011 $$ is the base, representing the growth factor per year. The exponent $$ (t-2018) $$ shows how many years have passed since 2018. This exponential function helps predict future population sizes based on current data and growth rates.Understanding exponential functions is essential because many real-world phenomena, like population growth, radioactive decay, and interest rates, follow exponential patterns. Mastering these concepts can help you solve various problems and make sense of different scenarios.