9. Suppose that the population of a city is 10,000 in 1970 and 50,000 in 1990. Assume further that the population is given by the formula

A = Pert

where P is the population in the year 1970, t is the number of years since 1970, and r is a suitable constant.

First substitute the values for 1970 into the equation. In 1970, t = 0, and A = 10,000

10,000 = Per(0)

10,000 = Pe0

But anything to the 0th is 1 so we get

10,000 = P

The initial amount is the y intercept.

NOw that we know P, we can substitute our values for 1990 into the equation. In 1990, t = 20 and P = 50,000

50,000 = 10,000er(20)

Divide both sides by 10,000

5 = e20r

Take natural logs of both sides

ln 5 = 20r

Divide both sides by 20

ln 5 is approximately 1.6094, so the answer is approximately

r = 0.08047

This gives us our equation

A = 10,000e0.08047t

b) With this formula we can predict the population in 2000. In 2000, t will be 30

A = 10,000e0.08047(30)

A = 10,000e2.414

A = 111,785

Since we are using decimal approximations, you will probably get a slilghtly different answer if you round off differently than was done here. Our calculations used numbers which were rounded off to 4 significant digits, so the correct answer should be

111,800

when rounded to 4 significant digits or the nearest hundred people.

c) When will the population reach 100,000?

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As you can see from the previous part of the problem the population should reach 100,000 before the year 2000. To find out exactly when, we use the formula

A = 10,000e0.08047t

and let A = 100,000

100,000 = 10,000e0.08047t

and solve for t. Divide both sides by 10,000

10 = e0.08047t

Take natural logs of both sides.

ln 10 = 0.08047t

Divide both sides by 0.08047

t = (ln 10)/0.08047

ln 10 is approximately 2.303 when rounded to 4 significant digits, so the answer is approximately

t = 28.62

28.62 years after 1970 would be 62% of the way through 1998.

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