4.3: Exponential Functions

India is the second most populous country in the world, with a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year5. We might ask if we can find a formula to model the population, P, as a function of time, t, in years after 2008, if the population continues to grow at this rate.

In linear growth, we had a constant rate of change – a constant number that the output increased for each increase in input. For example, in the equation f (x)=3x+4 , the slope tells us the output increases by three each time the input increases by one.

This population scenario is different – we have a percent rate of change rather than a constant number of people as our rate of change.

To see the significance of this difference consider these two companies:

• Company A has 100 stores, and expands by opening 50 new stores a year
• Company B has 100 stores, and expands by increasing the number of stores by 50% of their total each year.

Looking at a few years of growth for these companies:

Year	Stores, company	Stores, company	Starting with 100 each
0	100	100	Starting with 100 each
1	100 + 50 = 150	100 + 50% of 100 100 + 0.50(100) = 150	They both grow by 50 stores in the first year.
2	150 + 50 = 200	150 + 50% of 150 150 + 0.50(150) = 225	Store A grows by 50, Store B grows by 75
3	200 + 50 = 250	225 + 50% of 225 225 + 0.50(225) = 337.5	Store A grows by 50, Store B grows by 112.5

Notice that with the percent growth, each year the company grows by 50% of the current year’s total, so as the company grows larger, the number of stores added in a year grows as well. To try to simplify the calculations, notice that after 1 year the number of stores for company B was: 100 + 0.50(100) or equivalently by factoring 100(1+ 0.50) =150. We can think of this as “the new number of stores is the original 100% plus another 50%”.

After 2 years, the number of stores was:

• 150 + 0.50(150) or equivalently by factoring
• 150(1 + 0.50) now recall the 150 came from 100(1 + 0.50). Substituting that,
• 100(1 + 0.50)(1 + 0.50) = 100(1 + 0.50)2 = 225

After 3 years, the number of stores was:

• 225 + 0.50(225) or equivalently by factoring
• 225(1+ 0.50) now recall the 225 came from 100(1 + 0.50)2. Substituting that, 100(1+0.50)2(1+0.50)=100(1+0.50)3=337.5

From this, we can generalize, noticing that to show a 50% increase, each year we multiply by a factor of (1+0.50), so after n years, our equation would be:

• B(n) = 100(1+0.50)n

In this equation, the 100 represented the initial quantity, and the 0.50 was the percent growth rate. Generalizing further, we arrive at the general form of exponential functions.

Exponential Function

An exponential growth or decay function is a function that grows or shrinks at a constant percent growth rate. An exponential function has the form \(f(x)=a b^x\) where \(a\) is the initial or starting value of the function \(b\) is the growth factor (or decay factor)

• If \(b>1\), then the function is increasing (growth).
• If \(0<b<1\), then the function is decreasing (decay).

If \(r\) is the percent growth or decay rate (in decimal form), the exponential function can be written as follows: Growth: \(f(x)=a(1+r)^x\) or \(\quad\) Decay: \(f(x)=a(1-r)^x\)

To see more clearly the difference between exponential and linear growth, compare the two tables and graphs below, which illustrate the growth of company A and B described above over a longer time frame if the growth patterns were to continue.

Years	Company	Company
2	200	225
4	300	506
6	400	1139
8	500	2563
10	600	5767

Example \(\PageIndex{1}\)

Complete the table below.

Function	Initial Value	Growth or Decay	Rate (as a %)
𝑓(𝑥)=430(1.15)^𝑥
𝑓(𝑥)=23(0.8)^𝑥
	1.87	Growth	5 %
	52	Decay	10%

Solution

Function	Initial Value	Growth or Decay	Rate (as a %)
𝑓(𝑥)=430(1.15)^𝑥	430	Growth	1.15 – 1 = 0.15 Growth Rate: 15%
𝑓(𝑥)=23(0.8)^𝑥	23	Decay	1 – 0.8 = 0.2 Decay Rate = 20%
𝒇(𝒙)=𝟏.𝟖𝟕(𝟏+𝟎.𝟎𝟓)^𝒙 𝒇(𝒙)=𝟏.𝟖𝟕(𝟏.𝟎𝟓)^𝒙	1.87	Growth	5 %
𝒇(𝒙)=𝟓𝟐(𝟏−𝟎.𝟏𝟎)^𝒙 𝒇(𝒙)=𝟓𝟐(𝟎.𝟗)^𝒙	52	Decay	10%

Example \(\PageIndex{2}\)

T(q) represents the total number of Android smart phone contracts, in thousands, held by a certain Verizon store region measured q quarters since January 1, 2010.

Interpret all of the parts of the equation T(2) = 86(1.64)² = 231.3056.
Interpreting

Solution

Interpreting this from the basic exponential form, we know that 86 is our initial value. This means that on Jan. 1, 2010 this region had 86,000 Android smart phone contracts.

Since b = 1 + r = 1.64, we know that every quarter the number of smart phone contracts grows by 64%.

T(2) = 231.3056 means that in the 2nd quarter (or at the end of the second quarter) there were approximately 231,306 Android smart phone contracts.

You Try It \(\PageIndex{1}\)

a. Given the three statements below, identify which represent exponential functions.

1. The cost of living allowance for state employees increases salaries by 3.1% each year
2. The value of an investment is increasing by $300 each year
3. Tuition costs have increased by 2.8% each year for the last 3 years.

b. Looking at the following two equations that represent the balance in two different savings accounts, which account is growing faster, and which account will have a higher balance after 3 years?

A(t) = 1,000(1.05)^t B(t) = 900(1.075)^t

c. Complete the table. The first one has been done for you.

Function	Initial Value	Growth or Decay	Rate (as a %)
𝑓(𝑥)=615(0.79)^𝑥	615	Decay	21 %
𝑓(𝑥)=500(1.03)^𝑥
𝑓(𝑥)=3,200(1.2)^𝑥
𝑓(𝑥)=19,611(0.98)^𝑥
	50	Growth	12 %
	377	Decay	8 %

Answer

a. \(1 \& 3\) are exponential functions; they grow by a \(\%\) not a constant number.
b. \(B(t)\) is growing faster, but after 3 years \(A(t)\) still has a higher account balance
c.

\(f(x)=500(1.03)^x\)	500	Growth
(f(x)=3,200(1.2)^x\)	3,200	Growth
\(f(x)=19,611(0.98)^x\)	19,611	Decay
\(f(x)=50(1.12)^x\)	50	Growth
\(f(x)=377(0.92)^x\)	377	Decay

Example \(\PageIndex{3}\)

In the year 2008, India’s population was 1.14 billion and growing at a rate of 1.34% per year. Write an exponential function for India’s population, and use it to predict the population in 2020.

Solution

Using 2008 as our starting time \((t=0)\), our initial population will be 1.14 billion. Since the percent growth rate was \(1.34 \%\), our value for \(r\) is 0.0134 . Using the basic formula for exponential growth \(f(x)=a(1+r)^x\) we can write the formula,

\[
P(t)=1.14(1+0.0134)^t=1.14(1.0134)^t,
\]

where \(P(t)\) is the population of India (measured in billions) \(t\) years after 2008 .

To estimate the population in 2020, we evaluate the function at \(t=12\), since 2020 is 12 years after 2008 .

\[
P(12)=1.14(1.0134)^{12} \approx 1.337 \text { billion people in } 2020
\]

Example \(\PageIndex{2}\)

Bismuth-210 is an isotope that radioactively decays by about 13% each day, meaning 13% of the remaining Bismuth-210 transforms into another atom (polonium-210 in this case) each day. If you begin with 100 mg of Bismuth-210, how much remains after one week?

Solution

With radioactive decay, instead of the quantity increasing at a percent rate, the quantity is decreasing at a percent rate. Our initial quantity is \(a=100 \mathrm{mg}\), and our decay rate is \(13 \%\), so \(r=0.13\).

This gives the equation: \(Q(d)=100(1-0.013)^d=100(0.87)^d\) where \(Q(d)\) is the amount (in \(\mathrm{mg}\) ) of Bismuth-210 remaining after \(d\) days.

This can also be explained by recognizing that if \(13 \%\) decays, then \(87 \%\) remains.

After one week, 7 days, the quantity remaining would be \(Q(7)=100(0.87)^7=37.73 \mathrm{mg}\) of Bismuth-210 remains.

Example \(\PageIndex{3}\)

An investment is initially worth $10,000. Write a formula for the value of this investment for each situation described below.
a) The value increases by 5% every year.
b) The value increases by 5% every 3 years.
c) The value increases by 5% every 6 months.

Solution

In this problem, we are looking for three exponential functions of the form \(\mathrm{V}(t)=a b^t\), where \(\mathrm{V}(t)\) represents the value of the investment (in dollars) after \(t\) years. Since the initial value is \(\$ 10,000\), we know that \(a=10,000\). In each case, the value of the investment is increasing by \(5 \%\), so \(b=1+0.05=1.05\). The only difference in these three cases is the time period for which the \(5 \%\) increase occurs.

a) Since the \(5 \%\) increase occurs every year, we have \(\mathrm{V}(t)=10,000(1.05)^t\)

t	\(\mathrm{V}(t)=10,000(1.05)^t\)
0	\(10,000(1.05)^0 = 10,000\)
1	\(10,000(1.05)^1= 10,500\)
2	\(10,000(1.05)^2= 11,025\)
3	\(10,000(1.05)^3= 11,576\)

b) Since the \(5 \%\) increase occurs every three years, we need to multiply by the growth factor, 1.05 , only every three years. When \(t=3\) we multiply by the growth factor once. If \(t=6\), we multiply by the growth factor twice, and so on. The formula would then be \(\mathrm{V}(t)=10,000(1.05)^{t / 3}\)

t	\(\mathrm{V}(t)=10,000(1.05)^{t / 3}\)
0	\(10,000(1.05)^0/3 = 10,000(1.05)^0 = 10,000\)
3	\(10,000(1.05)^3/3= 10,000(1.05)^1 = 10,500\)
6	\(10,000(1.05)^6/3= 10,000(1.05)^2 =11,025\)
9	\(10,000(1.05)^9/3= 10,000(1.05)^3 =11,576\)

c) Since the \(5 \%\) increase occurs every six months, or twice each year, we need to multiply by the growth factor, 1.05 , two times each year. The formula would then be \(\mathrm{V}(t)=10.000(1.05)^{2 t}\)

t	\(\mathrm{V}(t)=10,000(1.05)^{2t }\)
0	\(10,000(1.05)^2(0) = 10,000(1.05)^0 = 10,000\)
1/2	\(10,000(1.05)^2(1/2)= 10,000(1.05)^1 = 10,500\)
1	\(10,000(1.05)^2(1)= 10,000(1.05)^2 =11,025\)
3/2	\(10,000(1.05)^2(3/2)= 10,000(1.05)^3 =11,576\)

To verify that these formulas are correct, we can compare the following tables of values

	a) Every year			b) Every 3 years			c) Every 6 months
t	\(\mathrm{V}(t)=10,000(1.05)^t\)		t	\(\mathrm{V}(t)=10,000(1.05)^{t / 3}\)		t	\(\mathrm{V}(t)=10,000(1.05)^{2t }\)
0	10,000		0	10,000		0	10,000
1	10,500		1	10164		1	10,500
2	11,025		2	10331		2	11,025
3	11,576		3	10,500		3	11,576
4	12,155		4	10,672		4	12,155
5	12,763		5	10,847		5	12,763
6	13,401		6	11,025		6	13,401

Example \(\PageIndex4}\)

Consider the following situations:

1. The population of a town is 83.6 thousand in the year 2005 and grows at a rate of 11.5% every 8 years.
2. A company’s sales are $14.2 million at the beginning of 2011, and have been decreasing at a rate of about 3.2% per quarter.

Write a formula for each of these situations using time, in years, as the input quantity

Solution

1. In this situation, let \(P(t)\) represent the population of the town (in thousands) \(t\) years after 2005. Since the population is increasing at a rate of \(14 \%\) every 8 years, the formula would then be:

\[
\begin{array}{c}
P(t)=83.6(1+0.115)^{t / 8} \\
P(t)=83.6(1.115)^{t / 8}
\end{array}
\]

2. In this situation, let \(S(t)\) represent the sales (in millions of dollars) \(t\) years after 2011 . Since sales are decreasing at a rate of \(3.2 \%\) per quarter, or 4 times each year, the formula for \(S(t)\) would then be:

\[
\begin{array}{c}
S(t)=14.2(1-0.032)^{4 t} \\
S(t)=14.2(0.968)^{4 t}
\end{array}
\]

You Try It \(\PageIndex{2}\)

An investment is initially worth $84,300. Write a formula for the function V(t), representing the value of this investment after t years in each of the following situations.

a) The value increases by 6.5% every year.
b) The value decreases by 3% every year.
c) The value increases by $600 every year.
d) The value decreases by $1400 every year.
e) The value increases by 26% every 2 years.
f) The value decreases by 13% every month.

Answer

a V(t) = 84,300(1.065)^t
b. V(t) = 84,300(0.97)^t
c. V(t) = 84,300 + 600^t
d. V(t) = 84,300 – 1,400^t
e. V(t) = 84,300(1.26)^t/2
f. V(t) = 84,300(0.87)^12t