obj. 14 exponential growth and decay (presentation)

Obj. 14 Exp. Growth and Decay

Unit 4 Exponential and Logarithmic Functions

Concepts and Objectives

� Exponential Growth and Decay (Obj. #14)

� Solve problems involving exponential growth and

decay.

� Set up and solve exponential variation problems

using both common and natural logarithms.

� Find half-life and doubling times.

Exponential Growth and Decay

� The general equation of an exponential function is

y = abx

where a and b are constants.

� Exponential growth occurs when b > 1.

� Exponential decay occurs when 0 < b < 1.

� The constant a is usually the starting value and b is the

percentage by which a is increasing or decreasing.


� Example: In 2006, the population of a country was 70

million and growing at a rate of 1.9% per year.

Assuming the percentage growth rate remains constant,

express the population, P, of this country (in millions) as

a function of t, the number of years after 2006.

Starting quantity: 70

Growth rate: 1+0.019 = 1.019

Equation: P = 70(1.019)t


� Example: A population of fish (P) starts at 8000 fish in

the year 2005 and decreases by 5.8% per year (t). What

is the predicted fish population in 2010?

Starting quantity: 8000

Decay rate: 1 – 0.058 = 0.942

Equation: P = 8000(0.942)t

In 2010 (t = 5): P = 8000(0.942)5

= 5934 fish


� If the relationship is continuously growing or decaying,

the equation can be written

y = aekx,

where a and k are constants and e is the base of the

natural logarithm.

� Exponential growth occurs when k > 0

� Exponential decay occurs when k < 0


� Example: World poultry production was 77.2 million

tons in the year 2004 and increasing at a continuous rate

of 1.6% per year (t). Estimate world poultry production

(P) in the year 2010.

Starting quantity: 77.2

Growth rate: 0.016

Equation: P = 77.2e0.016t

After 6 years: P = 77.2e(0.016)(6)

≈ 84.98 million tons


� How are you supposed to know which equation to use?!?

� If you are given a formula, you don’t have to worry

about it—just use the formula they give you.

� If the problem uses the word “continuous” or

“continuously”, use the equation with e.

� If the rate isn’t given (and you are not solving for the

rate), it doesn’t matter which version you use.

� If you’re still not sure, use the general form of the

exponential equation.

Exponential Variation

� Example: In a bacteria culture, there were 2000 bacteria

on Tuesday. On Thursday, the number has increased to

4500. Predict the number of bacteria that will be in the

culture next Tuesday.

Our starting quantity is 2000, and our starting time is

Tuesday. Thursday is 2 days away. This means that

=22000 4500b

=2 4500

2000b

= =2.25 1.5b


� Example (cont.):

Thus, our equation is

After 7 days, there should be

( )=72000 1.5C

≈ 34172 colonies

( )= 2000 1.5tC


� Example (cont.):

If we had wanted to use the natural log equation (e), it

would have worked much the same way:

( )=

22000 4500

ke

=2 4500

2000

ke

=2ln ln2.25ke

=2 ln2.25k

= ≈ln2.25

0.40552

k

( )( )=

0.4055 72000C e

≈ 34172 colonies

Half-Life and Doubling Time

� Half-life refers to the length of time it takes for an

exponential decay to reach half of its starting quantity.

� Doubling time refers to the length of time it takes for an

exponential growth to reach double its starting quantity.

� Both of these problems are actually worked the same

way. Use the general equation unless the rate is

continuous; in that case, you would use the equation

with e. To find the half-life (or doubling time), let a = 1

and set the equation equal to ½ (or 2) and solve for t.


� Example: Find the half-life of

(a) tritium, which decays at a rate of 5.471% per year

Since our decay rate is 0.05471, b will be 1–0.05471, or

0.94529.

=1

0.945292

t

=log0.94529 log0.5t

=log0.94529 log0.5t

=log0.5

log0.94529t ≈ 12.32 years


� Example: Find the half-life of

(b) a radioactive substance which decays at a continuous

rate of 11% per minute.

“Continuous” means we must use the e equation.

Because this decays, our rate, 0.11, is negative.

−=

0.11 1

2

te

−=

0.11ln ln0.5te− =0.11 ln0.5t

=−

ln0.5

0.11t ≈ 6.30 minutes


� Example: If the half-life of 600 g of a radioactive

substance is 3 years, how much of the substance will be

present after 6 years?

Since the rate isn’t mentioned, it really doesn’t matter

which equation we use:

=3 1

2b

( )=

3 1

2

ke

= 3 0.5b=

ln0.5

3k

( ) =6

3600 0.5 150 g ( ) =

ln0.56

3600 150 ge

Homework

� College Algebra & Trigonometry

� Page 475: 6-18 (evens), 27, 28, 33-36

� HW: 8, 12, 18, 28, 34, 36

� Functions Modeling Change (handout)

� Page 123: 15, 18, 19

� HW: 15, 19c

obj. 14 exponential growth and decay (presentation)

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