3 Exponential and Logarithmic Functions & Models
Introducing exponential functions without calculus presents a significant challenge. The simplest
approach is as a short-hand notation for repeated multiplication: for instance
a
5
= a · a · a · a · a
analogous to how multiplication represents repeated addition
5a = a + a + a + a + a
The problem with this approach is that it doesn’t help you understand what should be meant by, say,
a
3/4
or a
√
2
: multiplying something by itself ‘
√
2 times’ sounds
8
insane!
To rigorously address this problem requires continuity and other ideas surrounding the foundations
of calculus which you’ll encounter in upper-division analysis; topics unsuitable for this course. In-
stead, we assume some familiarity with exponential functions via introductory calculus, where they
are unavoidable and offer two ways to introduce exponential functions and e via modelling.
3.1 The Natural Growth Model
A basic model for any variable quantity is that its rate of change be proportional to the quantity itself. This
idea necessarily needs some calculus; as a differential equation,
dy
dx
= ky
where k is a constant; if k > 0 this is the natural growth equation, if k < 0 the natural decay equa-
tion. This is commonly encountered when modelling population growth; an otherwise unconstrained
population seems like its growth rate should be proportional to its size (twice the people, twice the
babies. . . ). This model is hugely applicable, since population can refer to essentially any quantifiable
value: people, bacteria, money, reagents in a chemical/nuclear reaction, etc.
Example 3.1. The simplest natural growth equation has k = 1:
dy
dx
= y
If a point (x, y) lies on a solution curve, then the differential equation tells
us the direction of travel of the solution. We may visualize this by drawing
an arrow with slope
dy
dx
= y; the arrows are tangent to any solution.
9
You
should easily be able to sketch some other solution curves.
You should, of course, recognize the graph. . .
−2
−1
1
2
y
−2 −1 1 2
x
8
The same issue arises for multiplication: 3
√
2 =
√
2 +
√
2 +
√
2 is relatively easy to understand, but how would you
convince someone what π
√
2 means?
9
A similar approach is available for any first-order differential equation
dy
dx
= F(x, y): the equation defines its slope field
(arrows), to which solution curves must be tangent.
30
Definition 3.2. Let a > 0 be constant. The exponential function with base a is f (x) = a
x
.
Recall the exponential laws, which are very natural when x, y, r are positive integers:
a
x+y
= a
x
a
y
a
x−y
=
a
x
a
y
(a
x
)
r
= a
rx
These hold for all exponents, with the same continuity caveats we saw previously. For modelling,
the crucial property of exponential functions is that they have proportional derivative.
Theorem 3.3. The rate of change of f (x) = a
x
is proportional to f (x). Specifically,
f
′
(x) = lim
h→0
a
x+h
− a
x
h
= a
x
lim
h→0
a
h
−1
h
so that f (x) = a
x
satisfies the natural growth/decay equation
dy
dx
= ky with proportionality constant
k = f
′
(0) = lim
h→0
a
h
−1
h
Example 3.4. We estimate the proportionality constant k = lim
h→0
a
h
−1
h
to 3 d.p. using a calculator for
four values of h:
a 2 2.5 2.7 2.75 3 5
a
0.1
−1
0.1
0.718 0.960 1.044 1.065 1.161 1.746
a
0.01
−1
0.01
0.696 0.921 0.998 1.017 1.105 1.622
a
0.001
−1
0.001
0.693 0.917 0.994 1.012 1.099 1.611
a
0.0001
−1
0.0001
0.693 0.916 0.993 1.012 1.099 1.610
What is happening to the proportionality constant as a increases? As h decreases?
It appears as if there is a special number somewhere between 2.7 and 2.75 for which the proportion-
ality constant is precisely k = 1.
Definition 3.5. The value e = 2.71828 . . . is the unique real number such that lim
h→0
e
h
−1
h
= 1.
The natural
10
exponential function exp(x) = e
x
has derivative
d
dx
e
x
= e
x
.
10
Natural here means unavoidable: an old cliche suggests that if aliens were to land on Earth, they’d have to understand e
given the technology they’d require to get here. Of course they’d likely use a different symbol; ours comes from Leonhard
Euler around 1728. Like π and
√
2, the constant e is an irrational number: its decimal representation contains no repeating
pattern. There isn’t the same geeky fascination with memorizing the digits of e as there is with π, neither is there an ‘e-day’
(Feb 7
th
at 6:28 p.m. anyone?).
31
The function f (x) =
1
2
e
x
is plotted in Example 3.1. Of course there are many other solutions to the
natural growth equation
dy
dx
= y: for any constants c, k,
y = ce
kx
=⇒
dy
dx
=
d
dx
ce
kx
= cke
kx
= ky
In fact the converse also holds; for the details, take a differential equations course!
Theorem 3.6. The solutions to the natural growth equation
dy
dx
= ky are precisely the functions
y(x) = y
0
e
kx
= y
0
exp(kx)
where y
0
= y(0) is the initial value.
Example 3.7. A Petri dish contains a population P(t) of bacteria satisfying the natural growth equa-
tion
dP
dt
= 0.5P where time is measured in weeks from the start of the year.
If P( 0) = 100 bacteria, then P(t) = 100e
0.5t
. Specifically, at the
end of January (4
3
7
weeks) one expects there to be
P(
31
7
) = 100 exp
31
14
= 915 bacteria
Note that the exponential doesn’t return 915 exactly; this is only
an approximation. Models like this work best for large popula-
tions where integer rounding errors are of minimal concern.
0
200
400
600
800
1000
P
0 1 2 3 4 5
t
Compound Interest and the Discovery of e
The first description of e came in 1683 when Jacob Bernoulli tried to model the growth of money in a
hypothetical bank account. We give a modernized version of his approach.
Example 3.8. $1 is deposited in an account paying 100% interest per year (nice!). Bernoulli observed
that the money in the account at the end of the year depends on when the interest is paid.
• If the interest is paid once at the end of the year (this is called simple interest), you’ll have $2.
• If half the interest (50¢) is paid at six months, then the balance ($1.50) earns
1
2
· 1.50 = 75¢
interest for the rest of the year; you’ll finish the year with $2.25 in the account.
• If the interest is paid in four installments, we have the following table of transactions (data is
rounded to the nearest cent)
Date Interest Paid Balance
1
st
Jan — $1
1
st
Apr 25¢ $1.25
1
st
July
1
4
·1.25 = 31¢ $1.56
1
st
Oct
1
4
·1.56 = 39¢ $1.95
New Year
1
4
·1.95 = 49¢ $2.44
More succinctly, the year-end balance is
1 +
1
4
4
= $2.44.
32
• More generally, if the interest is paid over n equally spaced intervals, the account balance at the
end of the year would be $
1 +
1
n
n
. Here are a few examples rounded things to 5 d.p.
Frequency Balance after 1 year ($)
Every month
1 +
1
12
12
= 2.61304
Every day
1 +
1
365
365
= 2.71457
Every hour
1 +
1
8760
8760
= 2.71813
Every second
1 +
1
31536000
31536000
= 2.71828
As the frequency of payment increases, it appears as if the balance is increasing to $e. . .
In fact this is a theorem, though it requires significant work (beyond this class) to prove it:
e = lim
n→∞
1 +
1
n
n
and more generally e
x
= lim
n→∞
1 +
x
n
n
This again shows that e arises very naturally.
Simple, Monthly & Continuous Interest In finance, interest is typically computed in one of three
ways. In each case we describe the result of investing $1 at an annual interest rate of r% =
r
100
.
Simple interest You are paid
r
100
dollars at the end of the year. Your invested dollar becomes 1 +
r
100
dollars.
Monthly interest Each month you are paid
r
12
% of your current balance. This amounts to a balance
of (1 +
r
1200
)
12
dollars at year’s end. The period need not be monthly: if interest is paid in n
installments, the balance would be ( 1 +
r
100n
)
n
.
Continuous interest After t years (can be any fraction of a year!) your dollar-balance is
e
rt
100
= exp
rt
100
= lim
n→∞
1 +
rt
100n
n
Example 3.9. A bank account earns 6% annual interest paid monthly. To what simple annual interest
rate does this correspond? Would you perfer an account paying 6% continuously?
At the end of the year, $1 becomes
(
1 + 0.0612
)
12
= 1.005
12
≈ 1.06168 . . .
corresponding to a simple interest rate of 6.17%. By cobntrast, 6% continuous interest would result
in your dollar becoming e
6/100
≈ 1.06184, corresponding to a (marginally) higher simple interest rate
of 6.18%. You should prefer this, particularly if you have a lot of money to invest! The difference is
more noticable with an investment of $1000 over ten years:
1000 ×1.005
120
= $1819.40 versus 1000e
0.6
= $1822.12
33
There are several reasons for these varying approaches, not all of them consumer-friendly:
1. Simple interest is simple! It is easy to understand and compute, but hard to decide how or even
whether to compute interest for parts of a year.
2. Monthly interest fits with most paychecks, so is sensible for loans, particularly mortgages.
3. Continuous interest allows the balance of an account to be found easily at any time, even be-
tween interest payment dates. It is also much easier to apply mathematical analysis (calculus).
4. A company can make an interest rate appear higher (if a savings account) or lower (if a loan) by
choosing which way to quote an interest rate.
Example 3.10. A bank quotes you a loan with a continuously compounded interest rate of 7%. If
you borrow $100,000, then at the end of the year you’ll owe
100000e
0.07
= $107, 250.82
not the $107,000 you might have expected! This corresponds to a simple interest rate (one payment
at the end of the year) of 7.25%.
11
Exercises 3.1. 1. Draw a slope field for the natural decay equation
dy
dx
= −
1
3
y and use it to sketch
the solution curve with initial condition y(0) = 6. What is the function y(x) in this case?
2. Which of the following would you prefer for a savings account? Why?
• 5% interest paid continuously.
• 5.05% compounded monthly.
• 5.1% paid at the end of the year.
3. You invest $1000 in an account that pays 4% simple interest per year.
(a) How much money will you have after 5 years?
(b) If you close the account after 2 years and 3 months, the bank needs to decide how much
interest to credit you with. Do this is two ways (the answers will be different!):
i. Compute using the simple interest rate for 2.25 years ((1 +
r
100
)
2.25
).
ii. Suppose that interest is paid at 4% for all completed years and then at 4% paid
monthly for any completed months of an incomplete year. Find the balance of the
account at closing.
4. See if you can explain why the proportionality constant for
1
a
x
is negative that for a
x
: that is,
lim
h→0
(
1
a
)
h
−1
h
= −lim
h→0
a
h
−1
h
Try to find both an algebraic reason and a pictorial one.
5. Sketch the function f (x) = e
−x
2
. Where have you seen this before, and what uses does this
function have?
11
In the US, mortgage companies typically quote an interest rate which they use to compound monthly. For example,
if the quoted rate is 7%, then the effective annual (simple) interest rate is
1 +
0.07
12
12
− 1 = 7.229%. By law, this higher
effective APR must be quoted somewhere, though it is unlikely to be as prominently posted. ..
34
3.2 Logarithmic Functions
Since e > 1 > 0, the exponential satisfies the following properties:
lim
x→−∞
e
x
= 0, lim
x→∞
e
x
= ∞,
d
dx
e
x
= e
x
> 0
The exp : R → (0, ∞) is a continuous, increasing function with domain R and range (0, ∞). It is
therefore invertible.
Definition 3.11. The natural logarithm ln : (0, ∞) → R is the
inverse function to the natural exponential. That is,
• If x > 0, then e
ln x
= x;
• If y ∈ R, then ln e
y
= y.
−1
0
1
2
y
1 2 3 4 5 6 7 8
x
y = ln x
e
e
−1
e
2
ln 4
Since exp and ln are inverse functions, we can solve equations in the usual way: for instance,
e
3x+1
= 100 =⇒ 3x + 1 = ln 100 =⇒ x =
1
3
(ln 100 −1) ≈ 1.202
One of the great advantages of logarithms is that they allow every exponential function to be ex-
pressed in terms of the natural exponential: by the exponential laws,
a
x
= (e
ln a
)
x
= e
x ln a
This identity is crucial for interpreting and analyzing natural growth models.
Example 3.12. A population of rabbits doubles in size every 6 months. If there are 10 rabbits at the
start of the year, how many rabbits do we expect there to be after 9 months, and how rapidly is the
population increasing (births/month).
We are told that the population of rabbits after t months is
P(t) = 10 ·2
t/6
After 9 months the population will be approximately
P(9) = 10 · 2
3/2
= 20
√
2 ≈ 28.28 rabbits
Moreover,
0
10
20
30
P
0 2 4 6 8 10
t
d
dt
P(t) =
d
dt
10e
t
6
ln 2
=
10 ln 2
6
2
t/6
=⇒ P
′
(9) =
10 ln 2
6
2
3/2
≈ 3.27 rabbits/month
If you ask students this question, what do you expect to be the most common incorrect answers?
Why?
35
The Logarithm Laws and General Logarithms
The logarithm laws should be familiar; they follow immediately from the above definition and the
exponential laws (page 31)
e
ln x+ln y
= e
ln x
e
ln y
= xy = e
ln xy
=⇒ ln xy = ln x + ln y
Similarly ln
x
y
= ln x −ln y and ln x
r
= r ln x (∗)
These laws allow us to solve more general exponential equations.
2
x
= 5 =⇒ x ln 2 = ln 5 =⇒ x =
ln 5
ln 2
≈ 2.322
More generally, if a > 0 and a = 1, then the exponential function with base a is invertible:
y = f (x) = a
x
= e
x ln a
=⇒ ln y = x ln a =⇒ x =
ln y
ln a
=⇒ f
−1
(x) =
ln x
ln a
Definition 3.13. Given a > 0 and a = 1, the logarithm with base a is the function
log
a
x :=
ln x
ln a
As the inverse of the base a exponential function y = a
x
, the base a logarithm satisfies
• If x > 0, then a
log
a
x
= x;
• If y ∈ R, then log
a
a
y
= y.
The natural logarithm has base e. Unless the base is very simple (e.g. a = 2 or 10), we typically stick
to using the natural logarithm. On a calculator, the ‘log’ button means log
10
.
Exercises 3.2. 1. Find the solution to the equation 4
2−
√
x
= 10.
2. Find the value of x which satisfies the equation 4
6x
= 8. Your answer should not contain any
logarithms. . .
3. Over one year, find the continuous interest rate s% corresponding to a simple rate of 5%.
4. y = a
x
satisfies the natural growth equation
dy
dx
= ky; what is the value of k?
5. Verify the remaining logarithm laws (∗) .
6. By differentiating the expression e
ln x
= x, verify that
d
dx
ln x =
1
x
.
7. Sketch a graph of the functions f (x) = log
2
x and g(x) = log
0.5
x. How are they related? What
happens to the graph of log
a
x as a changes?
8. Logarithms were originally invented not for calculus but to simplify and multiply large num-
bers. In the pre-calculator era, it was common for students to carry a book of log tables for this
purpose. Look up a log table and investigate how to use it.
36
3.3 Modifying the Natural Growth Model
In this section we discuss several examples of exponential models motivated by real-world situations.
Remember that modelling always involves some guesswork and assumptions, which necessarily come
with trade-offs: simpler assumptions/models are easier to analyze, but tend to be less accurate.
Modelling is always a part of a feedback loop:
Data/theory suggest a model whose predictions are tested against real-world data, sug-
gesting changes/improvements to the model.
Applied mathematicians typically desire a ‘Goldilocks’ model: complicated enough to make accurate
predictions without being too complicated to use.
Newton’s Law of Cooling
A just-poured cup of coffee at 210°F is left outside when the air temperature is 50°F. It seems obvious
that the coffee will cool down slowly towards 50°F; but how?
To help decide how to model this, ask yourself some questions:
1. When should the rate of cooling be most rapid?
2. What happens to the rate of change in the long run (large time)?
3. Can you suggest a family of functions which behave in this manner?
Hopefully it seems reasonable to model this with a shifted exponential function, where the tempera-
ture T(t) of the coffee at time t satisfies
T(t) = 50 + 160e
−kt
for some positive constant k. This satisfy all our criteria:
• T(0) = 210°F.
• As t increases, e
−kt
decreases to zero, so T(t) decreases
towards 50°F.
0
50
100
150
200
T
0 3 6 9 12
t
• The rate of cooling
dT
dt
= 160ke
−kt
is largest at t = 0 and decreases as t increases.
To complete the model, it is enough to supply one further data point.
Suppose after 2 minutes that the temperature of the coffee is 140°F. How long does it take for the
coffee to cool to 100°F?
We know that 140 = T(2) = 50 + 160e
−2k
, whence
e
−2k
=
140 −50
160
=
9
16
=⇒ e
−k
=
3
4
=⇒ T(t) = 50 + 160
3
4
t
When T(t) = 100, we see that
3
4
t
=
100 −50
160
=
5
16
=⇒ t =
ln
5
16
ln
3
4
=
ln 16 − ln 5
ln 4 − ln 3
≈ 4.043 minutes
37
This is an example of a general model called Newton’s law of cooling, which asserts that the rate of
temperature change of a body is proportional to the difference between the body and its surroundings.
We have a simple modification of Theorem 3.6
Corollary 3.14. If M and k are constant, then
dy
dt
= k(M − y) ⇐⇒ y(t) = M + (y
0
− M)e
−kt
where y
0
= y(0) is the initial value.
The Logistic Model
The natural growth model has one enormous drawback when applied to real-world populations: if
k > 0, then a function y(t) = y
0
e
kt
grows unboundedly! In typical situations, environmental limitations
(availability of food, water, space) mean that populations do not explode like this. The logistic model
attempts to describe this phenomenon; it is based on two assumptions:
• When a population y is small, we want it to grow nat-
urally
dy
dt
∝ y.
• We want y to approach a positive value M as t → ∞.
Given constants k, M > 0, the logistic differential equation
dy
dt
= ky(M − y)
0
y
0
t
M
dy
dt
< 0
dy
dt
> 0
accomplishes both requirements. M is often referred to as the carrying capacity of the environment.
Theorem 3.15. If y
0
= y(0), then the solution to the logistic differential equation is
y(t) =
y
0
M
y
0
+ (M − y
0
) e
−kMt
You can check directly that this satisfies the differential equation just by differentiating, though it’s a
little ugly. If you’ve studied differential equations the method of separation of variables supplies the
converse.
Example 3.16. A brewer pitches 100 billion yeast cells into a starter wort with the goal of growing it
to 200 billion cells. After one hour, the wort contains 110 billion cells.
(a) How long must the brewer wait if we use a natural growth model?
(b) How long must the brewer wait if we use a logistic model where we also assume that the wort
contains enough sugar to grow 250 billion yeast cells?
Let P(t) be the yeast population in billions at time t hours. We therefore have P(0) = 100 and
P(1) = 110, and want to find t such that P(t) = 200.
38
(a) The model is
dP
dt
= ky, which has solution
P(t) = P
0
e
kt
= 100e
kt
Evaluating at t = 1 yields 1.1 = e
k
whence
P(t) = 100(1.1)
t
=⇒ t =
ln( 0.01P)
ln 1.1
≈ 7.27 hours
(b) The model is
dP
dt
= ky(250 − y), with solution
P(t) =
25000
100 + (250 −100)e
−250kt
=
500
2 + 3e
−250kt
0
100
200
300
P
0 3 6 9 12
t
Evaluating at t = 1 yields
110 =
500
2 + 3e
−250k
=⇒ e
250k
=
33
28
whence
P(t) =
500
2 + 3
28
33
t
=⇒ t =
ln 6
ln
33
28
≈ 10.91 hours
The logistic model is easily generalized.
Example 3.17. The population P(t) (in 1000s) of fish in a lake obeys the logistic equation
dP
dt
=
1
16
P(10 − P)
where t is measured in months. The first graph shows how
the population recovers over a year if it starts at 2500 fish.
Now suppose 1000 fish are ‘harvested’ from the lake each
month. The new model is then
dP
dt
=
1
16
P(10 − P) −1 = −
1
16
(P
2
−10P + 16)
= −
1
16
(P −2)(P − 8)
Substituting Q = P − 2, this is again logistic!
dQ
dt
=
1
16
Q(6 − Q)
0
2
4
6
8
10
P
0 3 6 9 12
t
0
2
4
6
8
10
P
0 3 6 9 12
t
Provided the initial population P(0) = Q(0) + 2 is greater than 2000 fish, we expect the population
to eventually stabilize at 8000 fish, though it takes a long time to get close to this if we start, as in the
second graph, with only a little over 2000 fish.
39
Public-health Interventions
A population of 10,000 people is exposed to a novel virus. The best scientific understanding is that
1% of the susceptible population per day contracts the virus, the effects of the illness last ten days,
after which a patient recovers and is immune from reinfection.
1. Model the evolution of the sick and immune populations over the next 120 days.
Let u(t), s(t) and i(t) represent the uninfected, sick and immune populations on day t. Then
u(t + 1) = 0.99u(t), u(0) = 10000 =⇒ u(t) = 10000 · 0.99
t
The sick population is the sum of the previous 10 days’ decrease in the at risk population:
s(t) =
(
u(0) − u(t) = 10000( 1 −0.99
t
) if t ≤ 10
u(t −10) − u(t) = 10000(0.99
−10
−1)0.99
t
= 1057 · 0.99
t
if t > 10
The immune population is the difference between these and the total population
i(t) = 10000 − u(t) − s(t) =
(
0 if t ≤ 10
10000 −11057 ·0.99
t
if t > 10
After 120 days, we have
u(120) = 2994, s(120) = 316, i(120) = 6690
2. Suppose 6000 vaccines are available. Discuss how these should be deployed. What should the
goal be? Discuss the following strategies; for simplicity, assume the vaccines are 100% effective
and work instantly.
(a) Use all vaccine doses immediately.
(b) Wait 30 days until some people are immune, then use all vaccines on the uninfected pop-
ulation.
(c) Vaccinate 100 uninfected people per day.
(d) Wait until there are 6000 uninfected people remaining, then vaccinate them all at once.
It is much easier to analyze this problem using a spreadsheet, though we can also do things an-
alytically. Here are graphs of what happens under a non-intervention scenario and the first three
vaccination campaigns.
40
Exercises 3.3. 1. A cup of coffee is left outside on a warm day when the surrounding temperature
is 90°F. Suppose the initial temperature of the coffee is 200°F and that its temperature after 2
minutes is 170°F. Find the temperature as a function of time.
2. Consider Corollary 3.14.
(a) Check that y(x) = M + (y
0
− M)e
−kt
satisfies the differential equation.
(b) A student believes that Theorem 3.6 is true. How would you convince them, in the simplest
possible way, that the only solution to y
′
= k(M − y) is as given in part (a)?
3. For both models in Example 3.16, what is the maximum growth rate of the yeast population,
and at what time does it occur?
4. In Example 3.17, suppose the initial population is 3000 fish and 1000 fish are harvested per
month, how long does it take for the population to recover to 6000 fish?
5. (a) If public health officials wanted to eradicate the virus on page 40 completely by using all
vaccine doses on one day, on which day should they act?
(b) Find the number of uninfected people as a function of time under the 100 per day for 60
days vaccination scenario.
41
