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 <

__Questions__:

**1.** The number of compact discs purchased each year is increasing exponentially.

The number *N* purchased, **in millions**, is:

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

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 ,

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)

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 - ).*