1.5

Exponential Functions

DefinitionAn exponential function is a function of the form

P(t) = P0at

where a > 0 and a , 1.

We callP0 the initial value of the exponential function

a the base of the exponential function

1.5

The Initial Value

The number P0 in an exponential function P(t) = P0at plays animportant role:

P(0) = P0a0 = P0 · 1 = P0

It is called the “initial value”. So the “initial value” P0 of theexponential function is:I the function value at t = 0, i.e. P0 = P(0)I the vertical intercept of the graph of the functionI the “initial . . . ” in a word problem

1.5

The Initial Value

The number P0 in an exponential function P(t) = P0at plays animportant role:

P(0)

= P0a0 = P0 · 1 = P0

It is called the “initial value”. So the “initial value” P0 of theexponential function is:I the function value at t = 0, i.e. P0 = P(0)I the vertical intercept of the graph of the functionI the “initial . . . ” in a word problem

1.5

The Initial Value

The number P0 in an exponential function P(t) = P0at plays animportant role:

P(0) = P0a0

= P0 · 1 = P0

It is called the “initial value”. So the “initial value” P0 of theexponential function is:I the function value at t = 0, i.e. P0 = P(0)I the vertical intercept of the graph of the functionI the “initial . . . ” in a word problem

1.5

The Initial Value

The number P0 in an exponential function P(t) = P0at plays animportant role:

P(0) = P0a0 = P0 · 1 = P0

It is called the “initial value”. So the “initial value” P0 of theexponential function is:I the function value at t = 0, i.e. P0 = P(0)I the vertical intercept of the graph of the functionI the “initial . . . ” in a word problem

1.5

The Initial Value

The number P0 in an exponential function P(t) = P0at plays animportant role:

P(0) = P0a0 = P0 · 1 = P0

It is called the “initial value”.

So the “initial value” P0 of theexponential function is:I the function value at t = 0, i.e. P0 = P(0)I the vertical intercept of the graph of the functionI the “initial . . . ” in a word problem

1.5

The Initial Value

The number P0 in an exponential function P(t) = P0at plays animportant role:

P(0) = P0a0 = P0 · 1 = P0

It is called the “initial value”. So the “initial value” P0 of theexponential function is:I the function value at t = 0, i.e. P0 = P(0)

I the vertical intercept of the graph of the functionI the “initial . . . ” in a word problem

1.5

The Initial Value

The number P0 in an exponential function P(t) = P0at plays animportant role:

P(0) = P0a0 = P0 · 1 = P0

It is called the “initial value”. So the “initial value” P0 of theexponential function is:I the function value at t = 0, i.e. P0 = P(0)I the vertical intercept of the graph of the function

I the “initial . . . ” in a word problem

1.5

The Initial Value

The number P0 in an exponential function P(t) = P0at plays animportant role:

P(0) = P0a0 = P0 · 1 = P0

It is called the “initial value”. So the “initial value” P0 of theexponential function is:I the function value at t = 0, i.e. P0 = P(0)I the vertical intercept of the graph of the functionI the “initial . . . ” in a word problem

1.5

The Base

To see the role the base a plays in an exponential functionP(t) = P0at consider the quotients:

P(1)P(0)

=P0a1

P0= a

P(2)P(1)

=P0a2

P0a1 = a

P(3)P(2)

=P0a3

P0a2 = aP(4)P(3)

=P0a4

P0a3 = a

More generally,

P(t2)

P(t1)

=P0at2

P0at1=

at2

at1= at2−t1

So the base a is the “relative rate of change”.

1.5

The Base

To see the role the base a plays in an exponential functionP(t) = P0at consider the quotients:

P(1)P(0)

=P0a1

P0= a

P(2)P(1)

=P0a2

P0a1 = a

P(3)P(2)

=P0a3

P0a2 = aP(4)P(3)

=P0a4

P0a3 = a

More generally,

P(t2)

P(t1)

=P0at2

P0at1=

at2

at1= at2−t1

So the base a is the “relative rate of change”.

1.5

The Base

To see the role the base a plays in an exponential functionP(t) = P0at consider the quotients:

P(1)P(0)

=P0a1

P0= a

P(2)P(1)

=P0a2

P0a1 = a

P(3)P(2)

=P0a3

P0a2 = aP(4)P(3)

=P0a4

P0a3 = a

More generally,

P(t2)

P(t1)

=P0at2

P0at1=

at2

at1= at2−t1

So the base a is the “relative rate of change”.

1.5

The Base

To see the role the base a plays in an exponential functionP(t) = P0at consider the quotients:

P(1)P(0)

=P0a1

P0= a

P(2)P(1)

=P0a2

P0a1 = a

P(3)P(2)

=P0a3

P0a2 = a

P(4)P(3)

=P0a4

P0a3 = a

More generally,

P(t2)

P(t1)

=P0at2

P0at1=

at2

at1= at2−t1

So the base a is the “relative rate of change”.

1.5

The Base

To see the role the base a plays in an exponential functionP(t) = P0at consider the quotients:

P(1)P(0)

=P0a1

P0= a

P(2)P(1)

=P0a2

P0a1 = a

P(3)P(2)

=P0a3

P0a2 = aP(4)P(3)

=P0a4

P0a3 = a

More generally,

P(t2)

P(t1)

=P0at2

P0at1=

at2

at1= at2−t1

So the base a is the “relative rate of change”.

1.5

The Base

To see the role the base a plays in an exponential functionP(t) = P0at consider the quotients:

P(1)P(0)

=P0a1

P0= a

P(2)P(1)

=P0a2

P0a1 = a

P(3)P(2)

=P0a3

P0a2 = aP(4)P(3)

=P0a4

P0a3 = a

More generally,

P(t2)

P(t1)

=P0at2

P0at1=

at2

at1= at2−t1

So the base a is the “relative rate of change”.

1.5

The Base

To see the role the base a plays in an exponential functionP(t) = P0at consider the quotients:

P(1)P(0)

=P0a1

P0= a

P(2)P(1)

=P0a2

P0a1 = a

P(3)P(2)

=P0a3

P0a2 = aP(4)P(3)

=P0a4

P0a3 = a

More generally,

P(t2)

P(t1)=

P0at2

P0at1

=at2

at1= at2−t1

So the base a is the “relative rate of change”.

1.5

The Base

To see the role the base a plays in an exponential functionP(t) = P0at consider the quotients:

P(1)P(0)

=P0a1

P0= a

P(2)P(1)

=P0a2

P0a1 = a

P(3)P(2)

=P0a3

P0a2 = aP(4)P(3)

=P0a4

P0a3 = a

More generally,

P(t2)

P(t1)=

P0at2

P0at1=

at2

at1= at2−t1

So the base a is the “relative rate of change”.

1.5

The Base

To see the role the base a plays in an exponential functionP(t) = P0at consider the quotients:

P(1)P(0)

=P0a1

P0= a

P(2)P(1)

=P0a2

P0a1 = a

P(3)P(2)

=P0a3

P0a2 = aP(4)P(3)

=P0a4

P0a3 = a

More generally,

P(t2)

P(t1)=

P0at2

P0at1=

at2

at1= at2−t1

So the base a is the “relative rate of change”.

1.5

The Graph of a Exponential Function

1.5

The Graph of a Family of Exponential Functions

1.5

Recognizing Data from an Exponential Function

Question: How can we know if a function, represented as atable, can be given by an exponential function?

Easy answer: For equally spaced t values, if the ratios ofconsecutive P values are constant, then P can be given by anexponential function.

General answer: For unequally spaced t values, one has tosolve for the base a using each pair of consecutive P values, andthe function can be given by an exponential function if the avalues found this way are constant.

1.5

Recognizing Data from an Exponential Function

Question: How can we know if a function, represented as atable, can be given by an exponential function?

Easy answer: For equally spaced t values, if the ratios ofconsecutive P values are constant, then P can be given by anexponential function.

General answer: For unequally spaced t values, one has tosolve for the base a using each pair of consecutive P values, andthe function can be given by an exponential function if the avalues found this way are constant.

1.5

Recognizing Data from an Exponential Function

Question: How can we know if a function, represented as atable, can be given by an exponential function?

Easy answer: For equally spaced t values, if the ratios ofconsecutive P values are constant, then P can be given by anexponential function.

General answer: For unequally spaced t values, one has tosolve for the base a using each pair of consecutive P values, andthe function can be given by an exponential function if the avalues found this way are constant.

1.5

ExampleCan the following table correspond to an exponential function?

t 0 1 2 3 4P 16 24 36 54 81

Solution: The t values happen to be equally spaced, so we onlyneed to compute the ratios of consecutive P values:

P(1)P(0)

=2416

=32

P(2)P(1)

=3624

=32

P(3)P(2)

=5436

=32

P(4)P(3)

=8154

=32

The ratios are constant. So P(t) can be given by a exponentialfunction. Indeed, we also know that P0 = P(0) = 16, so

P(t) = 16 · (3/2)t

1.5

ExampleCan the following table correspond to an exponential function?

t 0 1 2 3 4P 16 24 36 54 81

Solution: The t values happen to be equally spaced, so we onlyneed to compute the ratios of consecutive P values:

P(1)P(0)

=2416

=32

P(2)P(1)

=3624

=32

P(3)P(2)

=5436

=32

P(4)P(3)

=8154

=32

The ratios are constant. So P(t) can be given by a exponentialfunction. Indeed, we also know that P0 = P(0) = 16, so

P(t) = 16 · (3/2)t

1.5

ExampleCan the following table correspond to an exponential function?

t 0 1 2 3 4P 16 24 36 54 81

Solution: The t values happen to be equally spaced, so we onlyneed to compute the ratios of consecutive P values:

P(1)P(0)

=2416

=32

P(2)P(1)

=3624

=32

P(3)P(2)

=5436

=32

P(4)P(3)

=8154

=32

The ratios are constant. So P(t) can be given by a exponentialfunction. Indeed, we also know that P0 = P(0) = 16, so

P(t) = 16 · (3/2)t

1.5

ExampleCan the following table correspond to an exponential function?

t 0 1 2 3 4P 16 24 36 54 81

Solution: The t values happen to be equally spaced, so we onlyneed to compute the ratios of consecutive P values:

P(1)P(0)

=2416

=32

P(2)P(1)

=3624

=32

P(3)P(2)

=5436

=32

P(4)P(3)

=8154

=32

The ratios are constant.

So P(t) can be given by a exponentialfunction. Indeed, we also know that P0 = P(0) = 16, so

P(t) = 16 · (3/2)t

1.5

ExampleCan the following table correspond to an exponential function?

t 0 1 2 3 4P 16 24 36 54 81

Solution: The t values happen to be equally spaced, so we onlyneed to compute the ratios of consecutive P values:

P(1)P(0)

=2416

=32

P(2)P(1)

=3624

=32

P(3)P(2)

=5436

=32

P(4)P(3)

=8154

=32

The ratios are constant. So P(t) can be given by a exponentialfunction. Indeed, we also know that P0 = P(0) = 16, so

P(t) = 16 · (3/2)t

1.5

e: a Special BaseWhile any real number a > 0, a , 1 can be used as a base for anexponential function, one base is of special interest:

the Euler’sconstant

e ≈ 2.71828182845904523536 . . .

a irrational number which can be define as

limn→∞(1 + 1/n)n

or ∞∑n=0

1n!

among many other ways. We will see why it is so special inchapter 3 and chapter 5.

1.5

e: a Special BaseWhile any real number a > 0, a , 1 can be used as a base for anexponential function, one base is of special interest: the Euler’sconstant

e ≈ 2.71828182845904523536 . . .

a irrational number which can be define as

limn→∞(1 + 1/n)n

or ∞∑n=0

1n!

among many other ways. We will see why it is so special inchapter 3 and chapter 5.

1.5

e: a Special BaseWhile any real number a > 0, a , 1 can be used as a base for anexponential function, one base is of special interest: the Euler’sconstant

e ≈ 2.71828182845904523536 . . .

a irrational number

which can be define as

limn→∞(1 + 1/n)n

or ∞∑n=0

1n!

among many other ways. We will see why it is so special inchapter 3 and chapter 5.

1.5

e: a Special BaseWhile any real number a > 0, a , 1 can be used as a base for anexponential function, one base is of special interest: the Euler’sconstant

e ≈ 2.71828182845904523536 . . .

a irrational number which can be define as

limn→∞(1 + 1/n)n

or ∞∑n=0

1n!

among many other ways.

We will see why it is so special inchapter 3 and chapter 5.

1.5

e: a Special BaseWhile any real number a > 0, a , 1 can be used as a base for anexponential function, one base is of special interest: the Euler’sconstant

e ≈ 2.71828182845904523536 . . .

a irrational number which can be define as

limn→∞(1 + 1/n)n

or ∞∑n=0

1n!

among many other ways. We will see why it is so special inchapter 3 and chapter 5.

1.5

Different Forms of Exponential Functions

A general exponential growth/decay function can be written inthe form

P(t) = P0at

for a > 1 (exponential growth) and 0 < a < 1 (exponentialdecay). But it is common to use the following equivalent forms:

P(t) = P0(1 + r)t (a = 1 + r)

where r is called the percentage growth/decay rate

P(t) = P0ekt (a = ek)

where k is called the continuous growth/decay rate

1.5

Different Forms of Exponential Functions

A general exponential growth/decay function can be written inthe form

P(t) = P0at

for a > 1 (exponential growth) and 0 < a < 1 (exponentialdecay). But it is common to use the following equivalent forms:

P(t) = P0(1 + r)t (a = 1 + r)

where r is called the percentage growth/decay rate

P(t) = P0ekt (a = ek)

where k is called the continuous growth/decay rate

1.5

Different Forms of Exponential Functions

A general exponential growth/decay function can be written inthe form

P(t) = P0at

for a > 1 (exponential growth) and 0 < a < 1 (exponentialdecay). But it is common to use the following equivalent forms:

P(t) = P0(1 + r)t (a = 1 + r)

where r is called the percentage growth/decay rate

P(t) = P0ekt (a = ek)

where k is called the continuous growth/decay rate

1.5

ExampleThe population of a small town is 3500 in year 2000, and it isgrowing at an annual rate of 10%. Write down the formula ofthe population of the small town P as a function of t, thenumber of years since 2000.

Solution: the initial population is P0 = P(0) = 3500.Each year the population is 10% more than the previous year,i.e., 1.1 times of the previous year’s population.so the base is a = 1.1.Therefore

P(t) = P0at = 3500 · 1.1t

Comment: one can also use P(t) = P0(1 + r)t where r is thepercentage growth rate (section 1.7).

1.5

ExampleThe population of a small town is 3500 in year 2000, and it isgrowing at an annual rate of 10%. Write down the formula ofthe population of the small town P as a function of t, thenumber of years since 2000.

Solution: the initial population is P0 = P(0) = 3500.

Each year the population is 10% more than the previous year,i.e., 1.1 times of the previous year’s population.so the base is a = 1.1.Therefore

P(t) = P0at = 3500 · 1.1t

Comment: one can also use P(t) = P0(1 + r)t where r is thepercentage growth rate (section 1.7).

1.5

ExampleThe population of a small town is 3500 in year 2000, and it isgrowing at an annual rate of 10%. Write down the formula ofthe population of the small town P as a function of t, thenumber of years since 2000.

Solution: the initial population is P0 = P(0) = 3500.Each year the population is 10% more than the previous year,

i.e., 1.1 times of the previous year’s population.so the base is a = 1.1.Therefore

P(t) = P0at = 3500 · 1.1t

Comment: one can also use P(t) = P0(1 + r)t where r is thepercentage growth rate (section 1.7).

1.5

ExampleThe population of a small town is 3500 in year 2000, and it isgrowing at an annual rate of 10%. Write down the formula ofthe population of the small town P as a function of t, thenumber of years since 2000.

Solution: the initial population is P0 = P(0) = 3500.Each year the population is 10% more than the previous year,i.e., 1.1 times of the previous year’s population.

so the base is a = 1.1.Therefore

P(t) = P0at = 3500 · 1.1t

Comment: one can also use P(t) = P0(1 + r)t where r is thepercentage growth rate (section 1.7).

1.5

ExampleThe population of a small town is 3500 in year 2000, and it isgrowing at an annual rate of 10%. Write down the formula ofthe population of the small town P as a function of t, thenumber of years since 2000.

Solution: the initial population is P0 = P(0) = 3500.Each year the population is 10% more than the previous year,i.e., 1.1 times of the previous year’s population.so the base is a = 1.1.

ThereforeP(t) = P0at = 3500 · 1.1t

Comment: one can also use P(t) = P0(1 + r)t where r is thepercentage growth rate (section 1.7).

1.5

ExampleThe population of a small town is 3500 in year 2000, and it isgrowing at an annual rate of 10%. Write down the formula ofthe population of the small town P as a function of t, thenumber of years since 2000.

Solution: the initial population is P0 = P(0) = 3500.Each year the population is 10% more than the previous year,i.e., 1.1 times of the previous year’s population.so the base is a = 1.1.Therefore

P(t) = P0at = 3500 · 1.1t

Comment: one can also use P(t) = P0(1 + r)t where r is thepercentage growth rate (section 1.7).

1.5

ExampleThe population of a small town is 3500 in year 2000, and it isgrowing at an annual rate of 10%. Write down the formula ofthe population of the small town P as a function of t, thenumber of years since 2000.

Solution: the initial population is P0 = P(0) = 3500.Each year the population is 10% more than the previous year,i.e., 1.1 times of the previous year’s population.so the base is a = 1.1.Therefore

P(t) = P0at = 3500 · 1.1t

Comment: one can also use P(t) = P0(1 + r)t where r is thepercentage growth rate (section 1.7).

1.5

ExampleSuppose the population of the United States is growing at aconsistent annual rate of 5% for the next century. Calculate thepopulation of the United States 100 years later. (Currently thepopulation is slightly more than three hundred million)

Solution: Use the initial population 300 million. Each year thepopulation is 5% more than the previous year, i.e., 1.05 times of theprevious year’s population. So the base is a = 1.05. And therefore

P(t) = 300 · (1.05)t

where t is the number of years since today.So the population 100 years later is

P(100) = 300 · (1.05)100 ≈ 39, 450.37

I.e., close to 40 billion.

1.5

ExampleSuppose the population of the United States is growing at aconsistent annual rate of 5% for the next century. Calculate thepopulation of the United States 100 years later. (Currently thepopulation is slightly more than three hundred million)

Solution: Use the initial population 300 million.

Each year thepopulation is 5% more than the previous year, i.e., 1.05 times of theprevious year’s population. So the base is a = 1.05. And therefore

P(t) = 300 · (1.05)t

where t is the number of years since today.So the population 100 years later is

P(100) = 300 · (1.05)100 ≈ 39, 450.37

I.e., close to 40 billion.

1.5

ExampleSuppose the population of the United States is growing at aconsistent annual rate of 5% for the next century. Calculate thepopulation of the United States 100 years later. (Currently thepopulation is slightly more than three hundred million)

Solution: Use the initial population 300 million. Each year thepopulation is 5% more than the previous year,

i.e., 1.05 times of theprevious year’s population. So the base is a = 1.05. And therefore

P(t) = 300 · (1.05)t

where t is the number of years since today.So the population 100 years later is

P(100) = 300 · (1.05)100 ≈ 39, 450.37

I.e., close to 40 billion.

1.5

ExampleSuppose the population of the United States is growing at aconsistent annual rate of 5% for the next century. Calculate thepopulation of the United States 100 years later. (Currently thepopulation is slightly more than three hundred million)

Solution: Use the initial population 300 million. Each year thepopulation is 5% more than the previous year, i.e., 1.05 times of theprevious year’s population.

So the base is a = 1.05. And therefore

P(t) = 300 · (1.05)t

where t is the number of years since today.So the population 100 years later is

P(100) = 300 · (1.05)100 ≈ 39, 450.37

I.e., close to 40 billion.

1.5

ExampleSuppose the population of the United States is growing at aconsistent annual rate of 5% for the next century. Calculate thepopulation of the United States 100 years later. (Currently thepopulation is slightly more than three hundred million)

Solution: Use the initial population 300 million. Each year thepopulation is 5% more than the previous year, i.e., 1.05 times of theprevious year’s population. So the base is a = 1.05.

And therefore

P(t) = 300 · (1.05)t

where t is the number of years since today.So the population 100 years later is

P(100) = 300 · (1.05)100 ≈ 39, 450.37

I.e., close to 40 billion.

1.5

ExampleSuppose the population of the United States is growing at aconsistent annual rate of 5% for the next century. Calculate thepopulation of the United States 100 years later. (Currently thepopulation is slightly more than three hundred million)

Solution: Use the initial population 300 million. Each year thepopulation is 5% more than the previous year, i.e., 1.05 times of theprevious year’s population. So the base is a = 1.05. And therefore

P(t) = 300 · (1.05)t

where t is the number of years since today.

So the population 100 years later is

P(100) = 300 · (1.05)100 ≈ 39, 450.37

I.e., close to 40 billion.

1.5

ExampleSuppose the population of the United States is growing at aconsistent annual rate of 5% for the next century. Calculate thepopulation of the United States 100 years later. (Currently thepopulation is slightly more than three hundred million)

Solution: Use the initial population 300 million. Each year thepopulation is 5% more than the previous year, i.e., 1.05 times of theprevious year’s population. So the base is a = 1.05. And therefore

P(t) = 300 · (1.05)t

where t is the number of years since today.So the population 100 years later is

P(100) = 300 · (1.05)100 ≈ 39, 450.37

I.e., close to 40 billion.

1.5

ExampleThe number of registered users of a popular social mediawebsite grew exponentially from 5, 000 to 1, 000, 000 over thelast 5 years. Find the annual percentage growth rate over thisperiod.

Solution: We should use the model P(t) = P0(1 + r)t with P0 = 5000.We know that P(5) = 1000000, so we get the equation

1000000 = P(5) = 5000(1 + r)5

simplify, we get

200 = (1 + r)5 ⇒ 1 + r = 20015 ≈ 2.8854

and thusr ≈ 1.8854

I.e., the annual percentage growth rate is approximately 188.54%.

1.5

ExampleThe number of registered users of a popular social mediawebsite grew exponentially from 5, 000 to 1, 000, 000 over thelast 5 years. Find the annual percentage growth rate over thisperiod.

Solution: We should use the model P(t) = P0(1 + r)t

with P0 = 5000.We know that P(5) = 1000000, so we get the equation

1000000 = P(5) = 5000(1 + r)5

simplify, we get

200 = (1 + r)5 ⇒ 1 + r = 20015 ≈ 2.8854

and thusr ≈ 1.8854

I.e., the annual percentage growth rate is approximately 188.54%.

1.5

ExampleThe number of registered users of a popular social mediawebsite grew exponentially from 5, 000 to 1, 000, 000 over thelast 5 years. Find the annual percentage growth rate over thisperiod.

Solution: We should use the model P(t) = P0(1 + r)t with P0 = 5000.

We know that P(5) = 1000000, so we get the equation

1000000 = P(5) = 5000(1 + r)5

simplify, we get

200 = (1 + r)5 ⇒ 1 + r = 20015 ≈ 2.8854

and thusr ≈ 1.8854

I.e., the annual percentage growth rate is approximately 188.54%.

1.5

ExampleThe number of registered users of a popular social mediawebsite grew exponentially from 5, 000 to 1, 000, 000 over thelast 5 years. Find the annual percentage growth rate over thisperiod.

Solution: We should use the model P(t) = P0(1 + r)t with P0 = 5000.We know that P(5) = 1000000, so we get the equation

1000000 = P(5) = 5000(1 + r)5

simplify, we get

200 = (1 + r)5 ⇒ 1 + r = 20015 ≈ 2.8854

and thusr ≈ 1.8854

I.e., the annual percentage growth rate is approximately 188.54%.

1.5

ExampleThe number of registered users of a popular social mediawebsite grew exponentially from 5, 000 to 1, 000, 000 over thelast 5 years. Find the annual percentage growth rate over thisperiod.

Solution: We should use the model P(t) = P0(1 + r)t with P0 = 5000.We know that P(5) = 1000000, so we get the equation

1000000 = P(5) = 5000(1 + r)5

simplify, we get

200 = (1 + r)5

⇒ 1 + r = 20015 ≈ 2.8854

and thusr ≈ 1.8854

I.e., the annual percentage growth rate is approximately 188.54%.

1.5

ExampleThe number of registered users of a popular social mediawebsite grew exponentially from 5, 000 to 1, 000, 000 over thelast 5 years. Find the annual percentage growth rate over thisperiod.

Solution: We should use the model P(t) = P0(1 + r)t with P0 = 5000.We know that P(5) = 1000000, so we get the equation

1000000 = P(5) = 5000(1 + r)5

simplify, we get

200 = (1 + r)5 ⇒ 1 + r = 20015 ≈ 2.8854

and thusr ≈ 1.8854

I.e., the annual percentage growth rate is approximately 188.54%.

1.5

ExampleThe number of registered users of a popular social mediawebsite grew exponentially from 5, 000 to 1, 000, 000 over thelast 5 years. Find the annual percentage growth rate over thisperiod.

Solution: We should use the model P(t) = P0(1 + r)t with P0 = 5000.We know that P(5) = 1000000, so we get the equation

1000000 = P(5) = 5000(1 + r)5

simplify, we get

200 = (1 + r)5 ⇒ 1 + r = 20015 ≈ 2.8854

and thusr ≈ 1.8854

I.e., the annual percentage growth rate is approximately 188.54%.

1.5

ExampleThe number of registered users of a popular social mediawebsite grew exponentially from 5, 000 to 1, 000, 000 over thelast 5 years. Find the annual percentage growth rate over thisperiod.

Solution: We should use the model P(t) = P0(1 + r)t with P0 = 5000.We know that P(5) = 1000000, so we get the equation

1000000 = P(5) = 5000(1 + r)5

simplify, we get

200 = (1 + r)5 ⇒ 1 + r = 20015 ≈ 2.8854

and thusr ≈ 1.8854

I.e., the annual percentage growth rate is approximately 188.54%.