ten

Exponential and
logarithmic
functions
Assumed knowledge
The content of the modules:
• Functions II
• Introduction to differential calculus
• Applications of differentiation
• Integration.
Furthermore, knowledge of the index laws and logarithm laws is assumed. These are
covered in the TIMES module Indices and logarithms (Years 9–10) and briefly revised at
the beginning of this module.
Motivation
The greatest shortcoming of the human race is our inability to understand
the exponential function.
—Albert A. Bartlett
Our world involves phenomena and objects onmany different scales.
Repeated multiplication by 10 can rapidly transform a microscopically small number to
an astronomically large one. Multiplying by 10 a few times takes us immediately from
the scale of atoms and molecules to the scale of microbiology, insects, humans, cities,
continents, planets and beyond—fromscales that are imperceptibly small to scales that
are almost unfathomably vast. There are only 17 orders of magnitude between the size
of a single human cell and the size of our solar system.1 Understanding the functions
1 A typical human cell is between 10?5 and 10?4 metres. The radius of the solar system, out to the orbit of
Neptune, is roughly 4.5£1012 metres.
A guide for teachers – Years 11 and 12 • {5}
involved in such repeated multiplication — namely, exponential functions such as 10x
—is a useful step towards a grasp of these enormities.
Exponential functions, with all their properties of sudden growth and decay, arise in
many natural phenomena, from the growth of living cells to the expansion of animal
populations, to economic development, to radioactive decay. The quote from Professor
Bartlett at the start of this section was made in the context of human population growth.
The inverses of exponential functions—namely, logarithmic functions—occur prominently
in fields as diverse as acoustics and seismology.
To understand these natural processes of growth and decay, it is important, then, to understand
the properties of exponential and logarithmic functions.
In this module, we consider exponential and logarithmic functions from a pure mathematical
perspective. We will introduce the function y ? ex , which is a solution of the
differential equation d y
dx
? y. It is a function whose derivative is itself. In the module
Growth and decay, we will consider further applications and examples.
Themodule Indices and logarithms (Years 9–10) covered many properties of exponential
and logarithmic functions, including the index and logarithm laws. Now, having more
knowledge, we can build upon what we have learned, and investigate exponential and
logarithmic functions in terms of their rates of change, antiderivatives, graphs and more.
In particular, we can ask questions like: How fast does an exponential function grow? It
grows rapidly! But, with calculus, we can give a more precise answer.
A brief refresher
To jog your memory, we recall some basic definitions and rules for manipulating exponentials
and logarithms. For further details, we refer to the module Indices and logarithms
(Years 9–10).
Logarithms and exponentials are inverse operations. In particular, for a E 1,
x ? ay () y ? loga x.
The following index laws hold for any bases a,b E 0 and any real numbersm and n:
aman ? am?n am
an
? am?n
(am)n ? amn (ab)m ? ambm
³ a
b
´m
?
am
bm .
{6} • Exponential and logarithmic functions
Some simple consequences of the index laws are, for a E 0 and a positive integer n:
a0 ? 1 a
1n
? n
p
a a?1 ?
1
a
.
The following logarithm laws hold for any base a E 1, any positive x and y, and any real
number n:
loga 1 ? 0 loga a ? 1
loga(x y) ? loga x ?loga y loga
x
y
? loga x ?loga y
loga
1
x
? ?loga x loga(xn) ? n loga x.
Also, recall the change of base formula:
logb x ?
loga x
loga b
,
for any a,b E 1 and any positive x.
Two approaches
In this module, we will introduce two new functions ex and loge x. We will do this in two
different ways.
The first approach develops the topic in an investigatory fashion, starting from the question:
‘What is the derivative of 2x ?’ However, as we proceed, we will point out some
shortcomings of this approach.
Alternatively, we can begin from a definition of loge x as an integral, and then define ex
as its inverse. The story is then told in a completely different order.
The first approach is probably easier for most students to understand, but the second
approach is more concise and rigorous.
In general, when telling a mathematical story, there are various goals such as elegance,
rigour, practicality, generality and understandability. Sometimes these goals conflict,
and we have to compromise. Sometimes developing a subject in the most logically concise
way does not make for easy reading. Aswith any other subject, learning mathematics
from multiple perspectives leads to a deeper and more critical understanding.
A guide for teachers – Years 11 and 12 • {7}
Content
How fast does an exponential function grow?
We will attempt to find the derivatives of exponential functions, beginning with 2x . This
is quite a long story, eventually leading us to introduce the number e, the exponential
function ex , and the natural logarithm. But we will then be able to differentiate functions
of the form ax in general.
The derivative of 2x
We begin by attempting to find the derivative of f (x) ? 2x , which is graphed as follows.
y
x
0
1
y = 2x
Graph of f (x) ? 2x .
Examining this graph, we can immediately say something about the derivative f 0(x).
The graph of y ? 2x is always sloping upwards and convex down. For large negative x, it
is very flat but sloping upwards. As x increases, the graph slopes increasingly upwards; as
x increases past 0, the gradient rapidly increases and the graph becomes close to vertical.
Therefore, we expect f 0(x) to be:
• always positive
• increasing
• approaching 0 as x !?1
• rapidly increasing for x positive
• approaching1as x !1.
{8} • Exponential and logarithmic functions
In other words, we expect f 0(x) to behave just like . . . an exponential function. (We will
see eventually that f 0(x) ? loge 2 ¢ 2x .)
Let’s first attempt to compute f 0(0) from first principles:
f 0(0) ? lim
h!0
f (h)? f (0)
h
? lim
h!0
2h ?1
h
.
This is not an easy limit to compute exactly, but we can approximate it by substituting
values of h.
Approximating f 0(0) for the function f (x) ? 2x
h
2h ?1
h
(to 6 decimal places)
0.001 0.693387
0.0001 0.693171
0.00001 0.693150
0.000001 0.693147
0.0000001 0.693147
From the table above, it appears that
f 0(0) ? lim
h!0
2h ?1
h
¼ 0.693147.
Let us press on and attempt to compute f 0(x) for x in general:
f 0(x) ? lim
h!0
f (x ?h)? f (x)
h
? lim
h!0
2x?h ?2x
h
? lim
h!0
2x(2h ?1)
h
? 2x ¢ lim
h!0
2h ?1
h
.
A guide for teachers – Years 11 and 12 • {9}
In the last step, we are able to take the 2x through the limit sign, since it is independent
of h. Note that the final limit is exactly the expression we found for f 0(0). So we can now
express the derivative as
f 0(x) ? f 0(0) ¢ 2x
¼ 0.693147 ¢ 2x .
We have found that the derivative of 2x is a constant times itself, confirming our initial
expectations. (We will see later that this constant is loge 2.)
The derivative of ax
There is nothing special about the number 2 above. If we take any number a E 1 and
consider f (x) ? ax , the graph would have a similar shape to that of 2x , and we could
carry out similar computations for f 0(0) and f 0(x):
f 0(0) ? lim
h!0
f (h)? f (0)
h
f 0(x) ? lim
h!0
f (x ?h)? f (x)
h
? lim
h!0
ah ?1
h
, ? lim
h!0
ax?h ?ax
h
? ax ¢ lim
h!0
ah ?1
h
.
We have found that, for any a E 1, the function f (x) ? ax has a derivative which is a
constant multiple of itself:
f 0(x) ? ax ¢ lim
h!0
ah ?1
h
? f 0(0) ¢ ax .
However, we need to better understand the limit involved. Clearly, if we can choose the
value of a so that this limit is 1, then f 0(x) ? f (x) and so f is its own derivative.
The number e
Let’s ponder further this limit for f 0(0), the derivative at 0 of f (x) ? ax :
f 0(0) ? lim
h!0
ah ?1
h
.
We have already found that, when a ? 2, the limit is approximately 0.693147. We can do
the same calculations for other values of a, and find the approximate limit. We obtain
the following table.
{10} • Exponential and logarithmic functions
Approximating f 0(0) for functions f (x) ? ax
a lim
h!0
ah ?1
h
(to 6 decimal places)
1 0
2 0.693147
3 1.098612
4 1.386294
5 1.609438
It appears that, when a increases, the limit for f 0(0) also increases. This is not too difficult
to prove.
Exercise 1
Show that, if 1 · a C b and h E 0, then
ah ?1
h
C
bh ?1
h
,
and hence explain why
lim
h!0
ah ?1
h
· lim
h!0
bh ?1
h
.
Geometrically, thismeans that, as a increases, the graph of y ? ax becomes more sharply
vertical, and the gradient of the graph at x ? 0 increases.
y
x
0
1
y = 2x y = 3x
Graphs of y ? 2x and y ? 3x with tangents shown at x ? 0.
A guide for teachers – Years 11 and 12 • {11}
When a ? 2, we have a gradient at x ? 0 of 0.69315. When a ? 3, we have a gradient of
1.0986. So we expect that there is a single value of a, between 2 and 3, for which the
gradient at x ? 0 is 1. It turns out that there is such a number,2 which we shall call e. The
number e is approximately 2.718281828.
Thus, the function f (x) ? ex has f 0(0) ? 1 and, since f 0(x) ? f 0(0) ¢ ex , we have f 0(x) ? ex .
The function ex is its own derivative. Equivalently, in Leibniz notation, y ? ex satisfies
d y
dx
? ex or, equivalently,
d y
dx
? y.
The function f (x) ? ex is often called the exponential function, and sometimes written
as expx.
Note. This approach may appear to be a sleight of hand. We didn’t really ‘prove’ that the
derivative of ex is itself, we just defined e to make it true. But the key point is that there
is a number e that makes the function ex its own derivative. We have given an argument
(although not a rigorous proof) as to why there is such a number.
Summary
• We considered the derivative of f (x) ? 2x and found that
f 0(x) ? 2x ¢ lim
h!0
2h ?1
h
¼ 0.693147 ¢ 2x ,
so f 0(x) is a constant multiple of f (x).
• We considered the derivative of the general function f (x) ? ax , where a E 1, and
found that
f 0(x) ? ax ¢ lim
h!0
ah ?1
h
,
so f 0(x) is a constant multiple of f (x).
• We defined the number e so that the function f (x) ? ex is its own derivative, that is,
f 0(x) ? f (x).
2 However, note we have not shown this. We have not shown that the limit limh!0
ah?1
h is a continuous
function of a; otherwise, this limit might never take the value 1. And we have not shown that the limit is
a strictly increasing function of a; if not, there might be multiple values of a for which the limit is 1. The
alternative treatment later in this module avoids these issues.
{12} • Exponential and logarithmic functions
Example
Define f : R!R by f (x) ? ex2 . What is f 0(x)?
Solution
Use the chain rule. Let g (x) ? ex and h(x) ? x2, so that f (x) ? g (h(x)). Then
f 0(x) ? g 0(h(x)) ¢h0(x)
? 2x ex2
.
Exercise 2
Find the derivatives of the following functions:
a f (x) ? x2ex b f (x) ? eex .
We will next introduce the natural logarithm. We will then be able to better express
derivatives of exponential functions.
The natural logarithm
The logarithm to the base e is an important function. It is also known as the natural
logarithm. It is defined for all x E 0:
y ? loge x () x ? e y .
The alternative notation lnx (pronounced ‘ell-en’ x) is often used instead of loge x.
y
x
0 1