Exponential Growth or Decay Function

Let be the amount or number present at time t = 0. Then, under certain conditions, the amount present at any time t is given by , where k is a constant.

When k > 0, the function describes growth.

When k < 0, the function describes decay.

Example. A sample of 300 grams of plutonium 241 decays according to the function

where t is time in years. Find the amount of the sample remaining after (a) 4 years (b) 8 years (c) 20 years, then (d) find the half-life.

Solution.

(a) represents the initial amount, so = 300 grams. t represents time in years, so

t = 4:

	Given formula
	Substitute 4 for t and 300 for A
	Simplify
A(4) » 243

243 grams remain after 4 years.

(b) t = 8

	Given formula
	t = 8
	Simplify
A(8) ) » 196

196 grams remain after 8 years.

(c) t = 20

	Given formula
	t = 20
	Simplify
A(20) » 104

104 grams remain after 20 years.

(d) The half-life is the time it takes for half of a given amount to decay. Since the initial amount is 300 grams, the half-life is the time it takes for this amount to decay to 150 grams. So, we have = 300 and A(t) = 150:

	Given formula
150 = 300e	Substitution
	Divide both sides by 300
0.5 = e	Simpllify
ln 0.5 = ln e	Take ln of both sides
ln 0.5 = -0.053t	Property of logarithms
	Divide both sides by -0.053
	Simplify
t » 13

The half-life is approximately 13 years.

Example. A sample from a refuse deposit near the Strait of Magellan had 60% of the carbon 14 of a contemporary sample. How old was the sample?

Solution. The amount of carbon 14 present after t years is given by

where is the amount present in living plants and animals. In this problem, the amount of carbon 14 present is 60% of that in a living sample, or .

	Given formula
	Substitute for y.
	Divide both sides by y
	Simplify
	Take ln of both sides
	Property of logarithms
	Multiply both sides by 5700; divide both sides by -ln 2
t » 4201

The sample is approximately 4201 years old.

Continuous Compounding

If P dollars is deposited at a rate of interest r compounded continuously for t years, the final amount on deposit is A = Pe dollars.

Example. Suppose $12,000 is deposited in an account paying 4% interest compounded continuously for 7 years. Find the total amount on deposit at the end of 7 years.

Solution. P = 12,000 ; r = 0.04 (always change the rate to decimal form); t = 7

A = Pe	Given formula
A = 12000 × e	Substitution
A = 12000 × e	Simplify
A » 15,877.56

The total amount on deposit after 7 years is $15,877.56.

Example. How long will it take for $5000 to grow to $8400 at an interest rate of 6% if interest is compounded continuously?

Solution. P = 5000; A = 8400; r = 0.06

A = Pe	Given formula
8400 = 5000e	Substitution
	Divide both sides by 5000
1.68 = e	Simplify
ln 1.68 = ln e	Take ln of both sides
ln 1.68 = 0.06t	Property of logarithms
t » 8.6	Divide both sides by 0.06

It will take approximately 8.6 years for the investment to grow to $8400.