eleven

A rigorous approach to logarithms and exponentials
What does an exponential mean anyway?
Throughout this module, we’ve assumed that functions like f (x) ? 2x are defined for all
real numbers x. But are they really?
There’s no problem defining 2x when x is a positive integer; this just means repeated
multiplication and is certainly well defined.
There’s also no problem when x is a negative integer, using the index law
2?n ?
1
2n .
For example, 2?3 ? 1
8 .
Nor is there a problem when x is a rational number. Using the index laws again,
2
a
b ? (2a)
1
b ?
b p
2a.
For example, 2
32
?
p
23 ?
p
8 ? 2
p
2.
However, it’s not so clear what to do when x is irrational. What does 2
p
2 mean? So far in
the module, this issue has been quietly suppressed.
One approach we might use is continuity. We could take a sequence of rational numbers
r1, r2, r3, . . . which approach
p
2, and consider 2r1 ,2r2 ,2r3 , . . . . If these numbers approach
a limit, then we can call that limit 2
p
2.
In any case, it’s not a trivial matter to define exponential functions like 2x for irrational x.
One way to avoid all of the difficulties is to develop the entire story a different way, starting
with the logarithmic function. As discussed in theMotivation section, this approach
may appear less natural but is more rigorous and abstract.
A guide for teachers – Years 11 and 12 • {25}
The natural logarithm, rigorously
We begin by defining the natural logarithm as an integral.
Definition
For any real number x E 0,
lnx ?
Z x
1
1
t
dt .
This equation was exercise 11. It is now a definition.
y
0 1 x
ln x
t
1t
y =
Definition of the natural logarithm as an integral.
As an aside, note the standard fact that the integral of tn is
1
n ?1
tn?1.
This is true for any n 6? ?1. The integral above is of tn when n ? ?1, precisely the value of
n for which this standard formula does not apply.
From this definition, we can see immediately that ln1 ? 0. We can also see, using the
fundamental theorem of calculus, that the derivative of lnx is
1
x
. (We refer to the module
Integration for the details). So the function lnx is increasing, for all x E 0, as its gradient
1
x
is positive. (This can also be seen from the diagram above, where lnx is shown as a
signed area.) It now follows that lnx C 0, for x 2 (0, 1), and lnx E 0, for x 2 (1,1). It is
clear that lnx is a continuous function, and it’s not too difficult to show that lnx !?1
as x !1, and that lnx !?1 as x !0.
{26} • Exponential and logarithmic functions
Our newly defined function lnx and our previously defined function loge x both have the
same derivative
1
x
. Since ln1 ? 0 ? loge 1, it follows that lnx and loge x are in fact the
same function—that is, if you manage to overcome all the difficulties with our previous
definition of loge x and arrive at a well-defined function.
From the definition, it’s not clear that the new function lnx behaves like a logarithm at
all. However, we will now show directly that this new function obeys the logarithm laws.
We use a similar method to exercise 10.
Take a positive number y, considered as a constant, and differentiate the two functions
f (x) ? ln(x y) and g (x) ? lnx ?ln y. We obtain
f 0(x) ?
y
x y
?
1
x
and g 0(x) ?
1
x
,
so f 0(x) ? g 0(x), and therefore f (x), g (x) differ by a constant. As f (1) ? ln y ? g (1), it
follows that f (x) ? g (x). This shows that
ln(x y) ? lnx ?ln y,
and so proves one of the logarithm laws.
Using a similar method, we can show that
ln
³ x
y
´
? lnx ?ln y
and
ln
³ 1
x
´
? ?lnx.
Next, consider the two functions f (x) ? ln(xn) and g (x) ? n lnx, where n is any rational
number.3 Differentiating these functions gives
f 0(x) ?
1
xn
¢nxn?1 ?
n
x
and g 0(x) ?
n
x
.
As f and g have the same derivative, they must differ by a constant. We further have
f (1) ? 0 ? g (1), so f (x) ? g (x) and we have proved the logarithmlaw
ln(xn) ? n lnx.
As our new function lnx obeys the familiar logarithm laws, we are justified in calling it a
logarithm!
3 Recall we said before that we only really know how to take rational powers.
A guide for teachers – Years 11 and 12 • {27}
Exponentials, rigorously
Having established our new version of the natural logarithm function, we now turn to
exponentials.
The function lnx mapping (0,1) to R is a continuous function, strictly increasing from
?1 to ?1, and hence has an inverse function. This inverse function has domain R and
range (0,1). We will, for the moment, call this inverse function expx. So, by definition,
expx ? ln?1(x).
We then have
exp(lnx) ? x for all positive x,
ln(expx) ? x for all real x.
We define the number e to be exp1. So exp1 ? e and lne ? 1. It will turn out that
expx ? ex ; however this is not at all clear from the definition.
We can compute the derivative of expx, since it is the inverse of lnx, and we know that
the derivative of lnx is
1
x
. Let y ? expx. Then x ? ln y and we have
dx
d y
?
1
y
,
so
d
dx
expx ?
d y
dx
?
1
dx
d y
? y ? expx.
We can also show that exp satisfies the index laws. For instance, we have
ln
?
expx ¢exp y
¢
? ln(expx)?ln(exp y)
? x ? y
? ln
?
exp(x ? y)
¢
.
Here we just used a logarithm law and the fact that ln and exp are inverses. Since the
function ln is one-to-one and we have just shown that ln
?
expx ¢ exp y
¢
? ln
?
exp(x ? y)
¢
,
we conclude the index law
expx ¢exp y ? exp(x ? y).
Using a similar method, we can show that exp also obeys the index law
exp(x ? y) ?
expx
exp y
.
{28} • Exponential and logarithmic functions
For the remaining index law, take a rational number r ; we will show exp(r x) ? (expx)r .
We observe
ln
?
(expx)r ¢
? r ln(expx)
? r x
? ln
?
exp(r x)
¢
,
where we just used a logarithm law and the fact that ln and exp are inverses. We have
ln
?
(expx)r ¢
? ln
?
exp(r x)
¢
, and cancelling ln’s establishes the index law
(expx)r ? exp(r x),
for any rational number r . Since exp1 ? e, we then have, for any rational number r ,
expr ? exp(r ¢ 1) ? (exp1)r ? er .
We can now use this to define irrational powers. We declare for any real number x, possibly
irrational, that
ex ? expx.
We