NZ Level 7 (NZC) Level 2 (NCEA)
Applications of Exponential Functions (y=a^x, a^-x only)
Lesson

An exponential function is the appropriate model to use when a quantity is increasing or decreasing at a rate that depends on the quantity present.

For example, in the final rounds of a sporting competition, the number of competing teams is halved at every stage. Thus, if $16$16 teams reached the first semifinal, there would be $8$8 in the second semifinal, and so on. The number of teams playing drops from $16$16 to $8$8 and then to $4$4 and finally, $2$2 and we see that the reduction in the number of teams playing at each stage depends on the number in the previous round.

This is an example of exponential decay. The rate of decrease gets progressively smaller. Many processes show the opposite pattern and exhibit exponential growth. In this, the rate of increase increases progressively.

#### Example 1

Exponential growth is seen in the compound interest process. If a bank deposit accumulates interest at, say, $3.1%$3.1% per annum, an amount of $\$100$$100 would grow to \103.10$$103.10 in $1$1 year, but in the next years it would grow to $\$106.30$$106.30, \109.60$$109.60, $\$113.00$$113.00, \116.50$$116.50, and so on.

We see that the amount of increase is getting larger at each stage because the interest is calculated on a slightly larger principle each year: $\$3.10$$3.10, \3.20$$3.20, $\$3.30$$3.30, ... At first, the amount of interest is small but eventually, if the money stays in the account, the interest payments will become very large. Each year, the amount of the deposit is being multiplied by 1.0311.031. So, the amount AA of the deposit after nn years should be A=1.031^n\times100A=1.031n×100. In 2323 years, the initial deposit of \100$$100 will have grown to $A=1,031^{23}\times100=$A=1,03123×100= $\$201.00201.00.

#### Example 2

In a biological experiment, a sample of a microorganism was seen to double in size by cell division every $11$11 minutes.

We can work out how many cells there will be after several $11$11-minute periods by a step-by-step calculation if we know how many cells were present initially. Suppose there were $n$n cells present at the beginning of the experiment.

After $11$11 minutes there are $2n$2n cells. Then, after a further $11$11 minutes, there are $2\times2n$2×2n cells. After a third period of $11$11 minutes, there are $2\times2\times2n$2×2×2n cells. This pattern continues and we see that after $k$k groups of $11$11 minutes, there will be $2^kn$2kn cells.

The researcher observes that the increase occurs gradually and not in sudden bursts at the end of each $11$11 minutes. The number of minutes that have elapsed since the beginning of the experiment is $m=11k$m=11k. So, it seems reasonable to put $k=\frac{m}{11}$k=m11 and write down a formula that will give the number of cells $N$N after $m$m minutes, for any number of minutes, not just multiples of $11$11. This is

$N=2^{\frac{m}{11}}n$N=2m11n.

Suppose there were $5$5 cells present at the beginning of the experiment and the growth of the microorganism was observed for $100$100 minutes. How many cells would there be at the end of this time?

We have $n=5$n=5. The exponent $\frac{m}{11}$m11 is $\frac{100}{11}\approx9$100119. So, on substituting these numbers into the formula, we calculate $N=2^9\times5=2560$N=29×5=2560 cells.

#### Example 3

A certain radioactive isotope decays in such a way that after $175$175 years only half of the original quantity of the isotope remains.

Suppose $10$10 kg of the substance existed initially. After $175$175 years, only $5$5 kg will be left and then after a further $175$175 years, only $2.5$2.5 kg will be left, and so on. After $k$k groups of $175$175 years, we would expect $A=10\times\left(\frac{1}{2}\right)^k$A=10×(12)k of the substance to remain.

The number of years that have elapsed since the beginning of this experiment must be $y=175k$y=175k. So, we can put $k=\frac{y}{175}$k=y175 into the formula so that we no longer have to convert the time elapsed into groups of $175$175 years but can use just years instead.

Also, we can replace $\left(\frac{1}{2}\right)^k$(12)k with the equivalent expression  $2^{-k}$2k.

Thus, we arrive at a formula for the amount remaining after $y$y years:

$A=10\times2^{-\frac{y}{175}}$A=10×2y175

In $1000$1000 years there will be $A=10\times2^{-\frac{1000}{175}}=0.19$A=10×21000175=0.19 kg of the isotope remaining.

#### Worked Examples

##### Question 1

Consider the graph shown.

1. Which of the following relationships could be represented by the given graph?

The number of people ($y$y) attending a parent/teacher conference when there are $x$x parents, each bringing $2$2 children.

A

The number of layers ($y$y) resulting from a rectangular piece of paper being folded in half $x$x times.

B

The number of handshakes ($y$y) made by $x$x people in a room if every person shakes hands with every other person.

C

The number of people ($y$y) attending a parent/teacher conference when there are $x$x parents, each bringing $2$2 children.

A

The number of layers ($y$y) resulting from a rectangular piece of paper being folded in half $x$x times.

B

The number of handshakes ($y$y) made by $x$x people in a room if every person shakes hands with every other person.

C
2. The equation of the graph is $y=2^x$y=2x.

If a rectangular piece of paper of thickness $0.02$0.02 mm is folded $11$11 times, determine the total resulting thickness.

##### Question 2

An initial chemical reaction results in a chain of chemical reactions that follow.

The total number of reactions after $t$t seconds is given by the equation $y=4^t$y=4t.

1. How many chemical reactions have occurred after the $3$3rd second?

2. How many chemical reactions occur in the $7$7th second only?

3. How many chemical reactions occur in the $8$8th second only?

4. How would you describe the rate of increase of the number of reactions?

The number of reactions increases rapidly at first, but then settles at a constant rate over time.

A

The number of reactions is increasing at a decreasing rate.

B

The number of reactions is increasing at a constant rate.

C

The number of reactions is increasing at an increasing rate.

D

The number of reactions increases rapidly at first, but then settles at a constant rate over time.

A

The number of reactions is increasing at a decreasing rate.

B

The number of reactions is increasing at a constant rate.

C

The number of reactions is increasing at an increasing rate.

D

##### Question 3

In a knockout mixed martial arts tournament, the winner of each round progresses to the next round until there are only two players left.

The diagram shows the draw for the final, semi-final and quarter final rounds.

1. Complete the table of values for the total number of players, $p$p, that the competition can accommodate given a number of rounds $r$r.

Number of Rounds ($r$r) $3$3 $4$4 $5$5 $6$6
Number of players ($p$p) $8$8 $\editable{}$ $\editable{}$ $\editable{}$
2. Suppose that the mixed martial arts tournament organisers want to make sure that each round of the tournament has every spot filled.

For what values of $p$p can a tournament be formed?

Any multiple of $4$4.

A

Any multiple of $2$2.

B

Any power of $2$2.

C

Any even number.

D

Any multiple of $4$4.

A

Any multiple of $2$2.

B

Any power of $2$2.

C

Any even number.

D
3. Plot the points from the table of values on the number plane.

4. What is the equation relating $r$r and $p$p?

Give your answer in the form $p=\editable{}$p=.

5. The organisers of a tournament can fit in $9$9 rounds of play.

How many players can they accept into the tournament?

### Outcomes

#### M7-2

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

#### 91257

Apply graphical methods in solving problems