Question

a logistic growth model for world population, f(x), in billions, x years after 1974 is ( f(x) = \frac{12.57}{1 + 4.11e^{-0.026x}} ). according to this model, when will the world population be 8 billion? according to this model, the world population will be 8 billion in (square). (round to the nearest whole number as needed.)

Explanation:

Step1: Set up the equation

We know the function \( f(x)=\frac{12.57}{1 + 4.11e^{-0.026x}} \) and we want to find \( x \) when \( f(x)=8 \). So we set up the equation:
\( 8=\frac{12.57}{1 + 4.11e^{-0.026x}} \)

Step2: Cross - multiply

Multiply both sides of the equation by \( 1 + 4.11e^{-0.026x} \):
\( 8(1 + 4.11e^{-0.026x})=12.57 \)

Step3: Distribute the 8

\( 8+32.88e^{-0.026x}=12.57 \)

Step4: Subtract 8 from both sides

\( 32.88e^{-0.026x}=12.57 - 8=4.57 \)

Step5: Divide both sides by 32.88

\( e^{-0.026x}=\frac{4.57}{32.88}\approx0.13899 \)

Step6: Take the natural logarithm of both sides

\( \ln(e^{-0.026x})=\ln(0.13899) \)
Since \( \ln(e^{a}) = a \), we have:
\( - 0.026x=\ln(0.13899) \)

Step7: Solve for x

We know that \( \ln(0.13899)\approx - 2.097 \)
So \( x=\frac{-2.097}{-0.026}\approx80.65 \)
Rounding to the nearest whole number, \( x\approx81 \)

Answer:

81