Question

A) P(t) = ___B) What population do you predict for the year 2000?Predicted population in the year 2000= ___ million people C) What is the doubling time?Doubling time= ____years.

Answer 1

Given:

The number of people in 1980 is 15 million.

The number of people in 1990 is 60 million.

Let t be the number of years.

The difference between 1990 and 1980 is 10 years.

Consider the general exponential equation

$P(t)=ae^{bt}$

Substitute t=0 and P(0)=15, we get

$15=ae^{b(0)}$$a=15$

Substitute a=15 in the general equation, we get

$P(t)=15e^{bt}$

Substitute t=10 and P(10)=60, we get

$60=15e^{b(10)}$

$\frac{60}{15}=e^{10b}$

$4=e^{10b}$

Taking log on both sides, we get

$In(4)=10b$

$1.38629=10b$

$b=\frac{1.38629}{10}=0.138629$

$b=0.139$

Substitute b=0.139 and a=15 in the general equation, we get

$P(t)=15e^{0.139(t)}$

Hence the exponential equation is

$P(t)=15e^{0.139(t)}$

In the year 2000, t=20.

Substitute t=20 in P(t), we get

$P(20)=15e^{0.139(20)}$

$P(20)=15e^{2.78}$

$P(20)=241.78$

$P(20)=242$

In the year 2000, the predicted population is 242 million.

The doubling time is the time when the population is double.

Substitute P(t)=30 to find the doubling time.

$30=15e^{0.139(t)}$

$\frac{30}{15}=e^{0.139(t)}$

$2=e^{0.139(t)}$

Taking log on both sides, we get

$In(2)=0.139\mleft(t\mright)^{}$

$\frac{0.6931}{0.139}=t$$t=4.98$$t=5\text{ years}$

The doubling time is 5 years.