EXPONENTIAL GROWTH AND DECAY QUESTIONS

Questions on the Exponential Function f (x) = ac x .

Hint: remember the order of operations. Always do exponentiation first, then multiply!

Reminder: If 0 < c < 1, it's a decay function. If c > 1, it's a growth function.

Questions:

1. The number of compact discs purchased each year is increasing exponentially.
The number N purchased, in millions, is:

N (t) = 7.5(6) ½ t

where t = 0 corresponds to 1985 so t represents the number of years since 1985.

a) Is this a growth or a decay function? Justify your answer.

b) Find the number of compact discs sold in the years 1986, 1987, 1990, 1995, and 2000.

(solution)

2. Suppose we invest \$50,000.00 at 9% interest compounded annually.

a) find a function rule for the amount of money in the account after t years.
b) find the amount of money in the account at t = 0, t = 4, t = 8, and t = 10 years.

(solution)

3. The e-coli bacteria, commonly found in the bladder of human beings, has a doubling period
of 20 minutes. If we let 3000g of this bacteria grow for t minutes, the number of bacteria present will be N(t) = 3000(2)
t / 20 .

How many grams of the bacteria will be present after 10 minutes? 20 minutes? 1 hour?

(solution)

4. We know that ¼ of all aluminum cans distributed are recycled each year. A soft drink company distributes 250,000 cans. The number of recycled cans still in use after t years is

N(t) = 250 000(¼) t .

a) Is this a growth or a decay function? Justify your answer.
b) How many cans are still in use after 0 years? 1 year? 4 years? 10 years?

(solution)

5. Office machines depreciate by 25% each year.

a) write the rule of correspondence for V(t), the value (after t years), of an office machine that cost \$5,200.00 to buy.

b) Find the value of that machine after 2 years, 5 years, and 10 years of use.

(solution)

Solutions:

1. a) This is a growth function since c > 1

b) Since N(t) = 7.5(6) ½t , for 1986 set t = 1, for 1987 set t = 2 etc.
N(1) = 7.5(6) ½(1) = 18.4 million
N(2) = 7.5(6) ½(2) = 7.5(6) = 45 million
N(5) = 661.4 million, N(10) = 58, 320 million, and N(15) = 5, 142,752.7 million

2. a) The function rule is: A(t) = 50,000(1 + 0.09) t , with A in \$ and t in years.

b) A(0) = 50,000(1.09) 0 = \$50,000 ,
A(4) = 50,000(1.09) 4 = \$70,579.08
A(8) = 50,000(1.09) 8 = \$99, 628.13
A(10) = 50,000(1.09) 10 = \$118,368.18

3. Since N(t) = 3000(2) t/20 ,
N(10) = 3000(2) 10/20 = 4243 grams
N(20) = 3000(2) 20/20 = 6000
N(60) = 3000(2) 60/20 = 24,000

4. a) This is a decay function since c < 1.

b) N(0) = 250 000(¼) 0 = 250,000.
N(1) = 250 000(¼) 1 = 62, 500.
N(4) = 250 000(¼) 4 = 977.
N(10) = 250 000(¼) 10 = 0 cans left since the answer is < 1.

5. a) The rule of correspondence is: V(t) = 5200 (1 – 0.25) t = 5200 (0.75) t .

b) V(2) = 5200 (0.75) 2 = \$2925
V(5) = 5200 (0.75) 5 = \$1233.98
V(10) = 5200 (0.75) 10 = \$292.83

(all content of the MathRoom Lessons © Tammy the Tutor; 2004 - ).