7.06 Exponential models

Exponential growth and decay models arise in many real-world situations. Here is the base function, its appearance and some real-world examples:

Growth	Decay
Examples: Balance of an investment earning compound interest Population growth Growth of cells Spread of a disease in a pandemic	Examples: Population decline of a threatened species  Radioactive decay Cooling of a liquid  Depreciation of a car

The base model for growth and decay is $y=A_0b^x$y=A0​bx, where $A_0$A0​ is the initial amount. When the multiplier $b$b is larger than one, we are in a growth situation and when the multiplier $b$b is between zero and one we are in a decay situation.

The models above have been drawn for $x\ge0$x≥0, as most models will have a horizontal axis representing time and start at an initial time of zero. However, do read questions carefully as this may not always be the case.

Notation: The variables $y$y and $x$x may change to highlight the variables in the model. Function notation is common such as $P(t)=3000\left(1.05\right)^t$P(t)=3000(1.05)t, to indicate the population as a function of time. Another common notation is sequence notation, such as $H_n=3(0.65)^n$Hn​=3(0.65)n, where $n$n indicates the number of times the multiplier has been applied. For example $H_3$H3​ may indicate the rebound height of a ball after $3$3 bounces. Sequence notation is particularly common for situations where $n$n can only take on integer values.

Let's look at some examples more closely and how the formula arises in practical situations.

 

Growth applications

An amount of $\$5000$$5000 is invested at $6%$6%p.a. compounded annually. Let's create a table of the balance of the account and try to generalise the findings.

Increasing an amount $A$A by $6%$6%, we require $106%$106% of the amount which is equivalent to multiplying by $1.06$1.06. We can think of in any of the following ways: $A\left(106%\right)=A\left(1+\frac{6}{100}\right)=A\left(1+0.06\right)$A(106%)=A(1+6100​)=A(1+0.06).

The amount $6%$6% is referred to the rate of growth of the investment.

In our problem, the factor must be applied each successive year as illustrated in this table:

Year	Calculation	Balance	Pattern
$0$0	$5000$5000	$\$5000$$5000	$5000$5000
$1$1	$5000\left(1.06\right)$5000(1.06)	$\$5300$$5300	$5000\left(1.06\right)^1$5000(1.06)1
$2$2	$5000\left(1.06\right)\left(1.06\right)$5000(1.06)(1.06)	$\$5618$$5618	$5000\left(1.06\right)^2$5000(1.06)2
$3$3	$5000\left(1.06\right)\left(1.06\right)\left(1.06\right)$5000(1.06)(1.06)(1.06)	$\$5955.08$$5955.08	$5000\left(1.06\right)^3$5000(1.06)3
$4$4	$5000\left(1.06\right)\left(1.06\right)\left(1.06\right)\left(1.06\right)$5000(1.06)(1.06)(1.06)(1.06)	$\$6312.38$$6312.38	$5000\left(1.06\right)^4$5000(1.06)4

We can see the balance grows exponentially, the interest paid increases each year as the interest from the previous year becomes part of the current balance. The pattern becomes clear, after $n$n years the principal  amount of $\$5000$$5000 will have the future value given by $A_n=5000\left(1.06\right)^n$An​=5000(1.06)n. 

We can see our general form for exponential growth: $A_n=A_0b^n$An​=A0​bn. Where $A_n$An​ is the amount in the account at the end of the $n^{th}$nth year and $A_0$A0​ is the principal(or initial) amount. 

Given the initial amount, $A_0$A0​, and rate of increase, $r$r, another useful form of a growth model is:

$y=A_0\left(1+r\right)^x$y=A0​(1+r)x

 

Worked example

Example 1

Sarah buys a piece of artwork for $\$1500$$1500 that is expected to appreciate (increase in value) by $8%$8% each year.

a) Find a model for $V_n$Vn​, the value of the artwork after $n$n years.

Think: Identify the initial amount and rate of increase and use appropriate variables in the formula $y=A_0\left(1+r\right)^x$y=A0​(1+r)x

Do: $A_0=\$1500$A0​=$1500, $r=8%$r=8%, hence: $V_n=1500\left(1+\frac{8}{100}\right)^n$Vn​=1500(1+8100​)n or $V_n=1500\left(1.08\right)^n$Vn​=1500(1.08)n

b) Find the estimated value of the artwork in $6$6 years' time, to the nearest cent.

Substitute $n=6$n=6 into the model and round to $2$2 decimal places.

$V_6$V6​	$=$=	$1500\left(1.08\right)^6$1500(1.08)6
	$\approx$≈	$\$2380.31$$2380.31

 

Rate of increase

From an equation of the form $y=A_0b^x$y=A0​bx, we can find the percentage rate of increase in $y$y for each unit of $x$x by: 

$\text{Rate of increase}=\left(b-1\right)\times100%$Rate of increase=(b−1)×100%

 

Worked example

Example 2

The population of foxes, $P$P, in a specified area, $t$t years after observation began, is modelled by the equation: $P\left(t\right)=300\left(1.25\right)^t$P(t)=300(1.25)t.

a) How many foxes are there initially?

The initial population is given by the number out the front. 300 foxes. We could also confirm this by substituting $t=0$t=0 (for the initial time) into the equation: 

$P(0)$P(0)	$=$=	$300\left(1.25\right)^0$300(1.25)0
	$=$=	$300\times1$300×1
	$=$=	$300$300 foxes

b) Approximately how many foxes are there after $5$5 years?

Substitute $t=5$t=5 into the equation and round the final answer to the nearest fox.

$P(5)$P(5)	$=$=	$300\left(1.25\right)^5$300(1.25)5
	$\approx$≈	$916$916 foxes

c) At what rate does the model suggest the population is increasing at per year?

We are multiplying by $1.25$1.25, this is equivalent to $125%$125%, so we are increasing by $25%$25% each year. This can also be found by the formula:

$\text{Rate of increase}$Rate of increase	$=$=	$\left(b-1\right)\times100%$(b−1)×100%
	$=$=	$\left(1.25-1\right)\times100%$(1.25−1)×100%
	$=$=	$25%$25%

 

Practice questions

Question 1

Question 2

 

Decay applications

A particular sports convertible costs $\$48000$$48000 and its value depreciates ( which means to lose value) at a rate of $18%$18% per year.

To find the car's value after $1$1 year we can find $18%$18% of $\$48000$$48000 and then subtract it. Alternatively, after losing $18%$18% we want to know what the remaining $82%$82% is. So after $1$1 year the car will be worth:

$V_1$V1​	$=$=	$\$48000\times0.82$$48000×0.82
	$=$=	$\$39360$$39360

After the second year, the value of the car will a drop by $18%$18% of $\$39360$$39360.  We can apply the multiplier of $0.82$0.82 once again and find the value after $2$2 years:

$V_2$V2​	$=$=	$\$48000\times0.82\times0.82$$48000×0.82×0.82
	$=$=	$\$48000\left(0.82\right)^2$$48000(0.82)2
	$=$=	$\$32275.20$$32275.20

 

We can see the value of the car decreases exponentially, since as the value decreases the $18%$18% reduction becomes smaller - so the value decreases more slowly as times passes. The pattern is very similar to our exponential growth case but with a multiplier between zero and one. After $n$n years the initial car value of $\$48000$$48000 will have decreased to a future value given by $V_n=48000\left(0.82\right)^n$Vn​=48000(0.82)n. 

We can see our general form for exponential decay: $V_n=A_0b^n$Vn​=A0​bn. Where $V_n$Vn​ is the value at the end of the $n^{th}$nth year and $A_0$A0​ is the principal(or initial) amount. 

Given the initial amount, $A_0$A0​, and rate of decrease, $r$r, another useful form of a decay model is:

$y=A_0\left(1-r\right)^x$y=A0​(1−r)x

 

Worked example

Example 3

A combine harvester bought for a farm for $\$420000$$420000 and depreciates at a rate of $21%$21%.

a) Find a model for $V_n$Vn​, the value of the harvester after $n$n years.

Think: Identify the initial amount and rate of decrease and use appropriate variables in the formula $y=A_0\left(1-r\right)^x$y=A0​(1−r)x

Do: $A_0=\$420000$A0​=$420000, $r=21%$r=21%, hence: $V_n=420000\left(1-\frac{21}{100}\right)^n$Vn​=420000(1−21100​)n or $V_n=48000\left(0.79\right)^n$Vn​=48000(0.79)n

b) Find the estimated value of the harvester in $3$3 years' time, to the nearest cent.

Substitute $n=3$n=3 into the model and round to $2$2 decimal places.

$V_3$V3​	$=$=	$48000\left(0.79\right)^3$48000(0.79)3
	$\approx$≈	$\$207076.38$$207076.38

 

Rate of decrease

From an equation of the form $y=A_0b^x$y=A0​bx, we can find the percentage rate of decrease in $y$y for each unit of $x$x by: 

$\text{Rate of decrease}=\left(1-b\right)\times100%$Rate of decrease=(1−b)×100%

 

Worked example

Example 4

The population of Tasmanian devils is declining due to disease. A model for the population, $P$P, in a specified area, $t$t years after observation began, is modelled by the equation: $P\left(t\right)=800\left(0.85\right)^t$P(t)=800(0.85)t.

a) How many devils are there initially?

The initial population is given by the number out the front: 800 devils.

b) At what rate does the model suggest the population is decreasing at per year?

We are multiplying by $0.85$0.85, this is equivalent to $85%$85%, so we are decreasing by $15%$15% each year. This can also be found by the formula:

$\text{Rate of increase}$Rate of increase	$=$=	$\left(1-b\right)\times100%$(1−b)×100%
	$=$=	$\left(1-0.85\right)\times100%$(1−0.85)×100%
	$=$=	$15%$15%

 

Practice questions

Question 3

Question 4

 

Solving for the exponent

In many cases, we may wish to know when a population or investment will reach a certain level. In example $4$4 above, perhaps we want to know when the population in the area is predicted to reach a critical level of $200$200 remaining devils. This would be found by solving the equation $200=800\left(0.85\right)^t$200=800(0.85)t, if we divide both sides by $800$800: $\left(\frac{1}{4}=\left(0.85\right)^t\right)$(14​=(0.85)t), we can see that solving for $t$t is equivalent to finding how many times we multiply $0.85$0.85 by itself to get $\frac{1}{4}$14​. By trial and error we can see that $t$t must be between $8$8 and $9$9 since $0.85^8\approx0.272$0.858≈0.272 and $0.85^9\approx0.232$0.859≈0.232. But how could we find $t$t more accurately?

This chapter has a focus on creating and applying exponential models and we will use technology to solve problems of this type. We can use technology to solve this by graphing both sides of the equation and finding the $x$x-coordinate of the point of intersection. That is, graph $y_1=800\left(0.85\right)^t$y1​=800(0.85)t and $y_2=200$y2​=200. This will provide a nice visualisation of the problem. Use your knowledge of exponential functions and a practical domain to obtain an appropriate view window.

From the graph, we can see the model predicts the population will reach a critical level in approximately $8.53$8.53 years' time. We can also solve this using an equation solver function in the calculator. 

 

Practice question

Question 5

 

Finding equations in different forms

Depending on the information provided it may be more convenient to write a model in a particular form. 

Do we have enough information to write models for the following:

• Find a model for the value of an investment after $n$n years, given an initial investment of $\$2000$$2000 and a rate of $5%$5% p.a. compounded monthly.
• Starting with an initial amount of $300$300 g. Find a model for the amount of an isotope remaining after $t$t years if it has a half-life of $6$6 years. 
• Find a model for a population of bacteria after $t$t hours, which starts with $500$500 and triples every $4$4 hours.

 

Let's look at each case separately.

 

Worked examples

Example 5 

Find a model for the value of an investment after $n$n years, given an initial investment of $\$2000$$2000 and a rate of $5%$5% p.a. compounded monthly.

If we were given the monthly rate then the model for the value after $t$t months would be: $A\left(t\right)=2000\left(1+r\right)^t$A(t)=2000(1+r)t

However, we have the annual rate and want the value after each year. Hence, we need to change the rate to a monthly rate by dividing by $12$12 and for each year this multiplier would be applied $12$12 times. So our new model becomes: $A\left(n\right)=2000\left(1+\frac{0.05}{12}\right)^{12n}$A(n)=2000(1+0.0512​)12n, where $n$n is in years.

In general, a model for the future value, $A$A, of a principal investment of $P$P, after $n$n years at a rate of $r$r p.a. compounded $k$k times per year is:

$A=P\left(1+\frac{r}{k}\right)^{kn}$A=P(1+rk​)kn

 

Example 6

Starting with an initial amount of $300$300 g. Find a model for the amount of an isotope remaining after $t$t years if it has a half-life of $6$6 years. 

Half-life, as the name suggests, is the time taken for a quantity to reduce to half its original amount. The term is commonly used in physics to describe how stable or unstable atoms are. We want a model that is going to have a multiplier of $\frac{1}{2}$12​ every $6$6 years. To create this model we have an initial amount and multiplier but need to change the index so that it will increase by $1$1 every $6$6 years. Hence, make the index $\frac{t}{6}$t6​ where $t$t is time in years. So our model is:

$W=300\left(\frac{1}{2}\right)^{\frac{t}{6}}$W=300(12​)t6​

 

Example 7

Find a model for a population of bacteria after $t$t hours, which starts with $500$500 and triples every $4$4 hours.

Similar to example $6$6 above but this time our multiplier is $3$3 and this needs to apply every $4$4 hours. Hence, our model is:

$P(t)=500\left(3\right)^{\frac{t}{4}}$P(t)=500(3)t4​

 

Extracting rates for different time periods

The examples above gave us models we can use to find the future values but the rate per time period was obscured. For instance, in example $7$7 the population was tripling every $4$4 hours but what was the rate of change per hour? We can use our index laws to rewrite the model so that it is in the form $y=A_0b^x$y=A0​bx and then find the rate from the value of $b$b. Let's look at a couple of concrete examples.

Worked examples

Example 8

What is the hourly rate of change for the population of bacteria from example $7$7, where the model is given by: $P\left(t\right)=500\left(3\right)^{\frac{t}{4}}$P(t)=500(3)t4​

Using index laws we can rewrite the formula as follows:

$P\left(t\right)$P(t)	$=$=	$500\left(3\right)^{\frac{t}{4}}$500(3)t4​
	$=$=	$500\left(\left(3\right)^{\frac{1}{4}}\right)^t$500((3)14​)t
	$\approx$≈	$500\left(1.316\right)^t$500(1.316)t

Hence, the population is growing at an approximate rate of $\left(b-1\right)\times100%=31.6%$(b−1)×100%=31.6% per hour.

Example 9

Find the effective annual rate of the following two investments and hence, identify which one gives a higher return.

Option A: $5.2%$5.2% p.a. compounded monthly or Option B: $5.18%$5.18% p.a. compounded weekly

An effective annualised rate takes into account the additional compounding payments. For both of these models the effective annual rate will be slightly higher than the stated rate but without calculating these it is hard to tell which investment will have a higher effective return. While option A looks like a higher rate of Option B compounds more regularly.

The model for Option A is $A_n=P\left(1+\frac{0.052}{12}\right)^{12n}$An​=P(1+0.05212​)12n, which can be rewritten as:

$A_n$An​	$=$=	$P\left(1+\frac{0.052}{12}\right)^{12n}$P(1+0.05212​)12n
	$=$=	$P\left(\left(1+\frac{0.052}{12}\right)^{12}\right)^n$P((1+0.05212​)12)n
	$\approx$≈	$P\left(1.0533\right)^n$P(1.0533)n

 Hence, Option A has an effective rate of approximately $5.33%$5.33% p.a..

The model for Option B is $A_n=P\left(1+\frac{0.0518}{52}\right)^{52n}$An​=P(1+0.051852​)52n, which can be rewritten as:

$A_n$An​	$=$=	$P\left(1+\frac{0.0518}{12}\right)^{52n}$P(1+0.051812​)52n
	$=$=	$P\left(\left(1+\frac{0.0518}{52}\right)^{52}\right)^n$P((1+0.051852​)52)n
	$\approx$≈	$P\left(1.0531\right)^n$P(1.0531)n

 Hence, Option B has an effective rate of approximately $5.31%$5.31% p.a..

Thus, Option A has a slightly better effective annual rate.

Example 10

A population is decreasing according to the following model: $P\left(t\right)=5000\left(0.8\right)^t$P(t)=5000(0.8)t, where $t$t is in years. What is the monthly rate of change for the population?

Using our index laws we need to adjust the time frame to be in months, we do this as follows:

$P\left(t\right)$P(t)	$=$=	$5000\left(0.8\right)^t$5000(0.8)t
	$=$=	$5000\left(\left(0.8\right)^{\frac{1}{12}}\right)^{12t}$5000((0.8)112​)12t
	$\approx$≈	$5000\left(0.982\right)^{12t}$5000(0.982)12t

Hence, if $n$n is in months the model would be $P\left(t\right)=5000\left(0.982\right)^n$P(t)=5000(0.982)n, which shows the population is decreasing at approximately $\left(1-b\right)\times100%=1.8%$(1−b)×100%=1.8% per month.

 

Practice questions

Question 6

Question 7

Question 8