Question
A logistic growth model for world population, f(x), in billions, x years after 1965 is f left parenthesis x right parenthesis equals StartFraction 12.57 Over 1 plus 4.11 e Superscript negative 0.026 x EndFraction
. According to this model, when will the world population be 10 billion?
Question content area bottom
Part 1
According to this model, the world population will be 10 billion in
enter your response here.
(Round to the nearest whole number as needed.)
. According to this model, when will the world population be 10 billion?
Question content area bottom
Part 1
According to this model, the world population will be 10 billion in
enter your response here.
(Round to the nearest whole number as needed.)
Answers
GPT-4o mini
To determine when the world population will reach 10 billion according to the logistic growth model provided, we need to solve the equation:
\[
f(x) = 10
\]
where
\[
f(x) = \frac{12.57}{1 + 4.11 e^{-0.026x}}.
\]
Setting the function equal to 10 billion gives us:
\[
\frac{12.57}{1 + 4.11 e^{-0.026x}} = 10.
\]
Now, we will solve for \( x \):
1. Multiply both sides by \( 1 + 4.11 e^{-0.026x} \):
\[
12.57 = 10(1 + 4.11 e^{-0.026x}).
\]
2. Distributing the 10:
\[
12.57 = 10 + 41.1 e^{-0.026x}.
\]
3. Subtract 10 from both sides:
\[
2.57 = 41.1 e^{-0.026x}.
\]
4. Divide both sides by 41.1:
\[
e^{-0.026x} = \frac{2.57}{41.1}.
\]
5. Calculate \( \frac{2.57}{41.1} \):
\[
e^{-0.026x} \approx 0.0625.
\]
6. Take the natural logarithm of both sides:
\[
-0.026x = \ln(0.0625).
\]
7. Solve for \( x \):
\[
x = \frac{\ln(0.0625)}{-0.026}.
\]
8. Calculate \( \ln(0.0625) \):
\[
\ln(0.0625) \approx -2.772588.
\]
9. Substitute this value:
\[
x = \frac{-2.772588}{-0.026} \approx 106.7.
\]
Since we want to round to the nearest whole number, we get:
\[
x \approx 107.
\]
Now, since \( x \) represents the number of years after 1965, we will add this to 1965 to find the year:
\[
1965 + 107 \approx 2072.
\]
Thus, according to this model, the world population will reach 10 billion in the year **2072**.
\[
f(x) = 10
\]
where
\[
f(x) = \frac{12.57}{1 + 4.11 e^{-0.026x}}.
\]
Setting the function equal to 10 billion gives us:
\[
\frac{12.57}{1 + 4.11 e^{-0.026x}} = 10.
\]
Now, we will solve for \( x \):
1. Multiply both sides by \( 1 + 4.11 e^{-0.026x} \):
\[
12.57 = 10(1 + 4.11 e^{-0.026x}).
\]
2. Distributing the 10:
\[
12.57 = 10 + 41.1 e^{-0.026x}.
\]
3. Subtract 10 from both sides:
\[
2.57 = 41.1 e^{-0.026x}.
\]
4. Divide both sides by 41.1:
\[
e^{-0.026x} = \frac{2.57}{41.1}.
\]
5. Calculate \( \frac{2.57}{41.1} \):
\[
e^{-0.026x} \approx 0.0625.
\]
6. Take the natural logarithm of both sides:
\[
-0.026x = \ln(0.0625).
\]
7. Solve for \( x \):
\[
x = \frac{\ln(0.0625)}{-0.026}.
\]
8. Calculate \( \ln(0.0625) \):
\[
\ln(0.0625) \approx -2.772588.
\]
9. Substitute this value:
\[
x = \frac{-2.772588}{-0.026} \approx 106.7.
\]
Since we want to round to the nearest whole number, we get:
\[
x \approx 107.
\]
Now, since \( x \) represents the number of years after 1965, we will add this to 1965 to find the year:
\[
1965 + 107 \approx 2072.
\]
Thus, according to this model, the world population will reach 10 billion in the year **2072**.