Math 113 Lecture #16
Appendix G: The Logarithm Defined as an Integral
The Natural Logarithm. You might recall that the Fundamental Theorem of Calculus Part I states that the antiderivative of a function continuous on [a, b] is the function
Z x
f (t) dt
a
which is continuous on [a, b] and differentiable on (a, b).
You might recall that the natural logarithmic function was defined as the inverse of the
exponential function.
However, we will use the FTC Part I to define the natural logarithmic function, and
define its inverse to be the exponential function.
We will show that all of familiar properties of logarithms and exponentials follow from
this FTC Part I definition of the natural logarithm function.
Here is the definition of the natural logarithmic function:
Z x
1
ln x =
dt.
1 t
The domain of ln x is x > 0 since y = 1/t is continuous (hence integrable) between 1 and
x for all x > 0.
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2
1
0
0
1
2
3
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5
t
Geometrically, the natural logarithmic function is the area under the hyperbola y = 1/x
from 1 to x:
Z x
Z x
Z 1
1
1
1
ln x =
dt > 0 for x > 1, and ln x =
dt = −
dt < 0 for 0 < x < 1.
1 t
1 t
x t
Clearly
Z
ln 1 =
1
1
1
dt = 0.
t
Furthermore, we know that
d
d
(ln x) =
dx
dx
Z
1
x
1
1
dt = .
t
x
Theorem: Laws of the Natural Logarithm. If x and y are positive numbers and r
is a rational number, then
x
= ln x − ln y, (3) ln(xr ) = r ln x.
(1) ln(xy) = ln x + ln y, (2) ln
y
Proof. (1) For a constant a (to be replaced by the variable y in a moment), let
f (x) = ln(ax).
By the Chain Rule,
1 d
a
1
(ax) =
= .
ax dx
ax
x
This means that f (x) and ln x have the same derivative, and so they differ by a constant:
f 0 (x) =
ln(ax) = f (x) = ln x + C.
We can solve for the value of the constant C by the choice of x = 1 (this because we
know ln 1 = 0):
ln a = ln 1 + C = 0 + C = C.
We now have
ln(ax) = ln x + ln a.
Replacing a by y gives Property (1):
ln(xy) = ln x + ln y.
To prove Property (2) we start with Property (1) applied to the case of x = 1/y:
ln
y
1
+ ln y = ln = ln 1 = 0.
y
y
This implies that
ln
1
= − ln y.
y
We use this and Property (1) to obtain Property (2):
x
1
1
ln = ln x
= ln x + ln = ln x − ln y.
y
y
y
Property (3) is a homework problem for your enjoyment.
As x → 0+ or x → ∞, the integral defining the natural logarithm becomes improper.
Are they convergent or divergent?
Theorem: Divergence of Natural Logarithm. lim+ ln x = −∞ and lim ln x = ∞.
x→∞
x→0
Remark. We might be tempted to do this:
Z
x
lim ln x = lim
x→∞
x→∞
1
1
dt = ∞
t
because the integral is a divergent improper integral, but the problem with this argument
is that we are assuming the limit we want to prove.
Proof. Instead, we make use of Property (3): for x = 2 and r = n a positive integer,
ln 2n = n ln 2.
Now ln 1 = 0 and since (d/dx) ln x = 1/x is positive, then ln x is increasing, so that
ln 2 > 0. Thus
lim ln 2n = lim n ln 2 = ln 2 lim n = ∞.
n→∞
n→∞
n→∞
Again since ln x is increasing, this limit holds for x as x → ∞ too.
If we now let s = 1/x then s → ∞ as x → 0+ , and so
1
lim+ ln x = lim ln
= − lim ln s = −∞.
s→∞
s→∞
x→0
s
This second limit also follows from a comparsion of the area under the graph of y = 1/t
from 1 to ∞ with that under the graph from 0 to 1.
We saw in this proof that ln x is increasing. It is also concave down, since we can compute
the second derivative:
d 1
1
d2
(ln x) =
= − 2 < 0.
2
dx
dx x
x
With this information we can sketch the graph of the natural logarithmic function.
2
1
x
0
1
2
3
4
5
0
!1
!2
!3
!4
!5
Now as ln 1 = 0 and lim ln x = ∞, and ln x is an increasing continuous function, there
x→∞
is by the Intermediate Value Theorem a unique value e for which
ln e = 1.
Just think, all of the properties of the natural logarithmic function are consequences of
Calculus!
The Exponential Function. Since the natural logarithmic function ln is increasing,
it is one-to-one, and so has an inverse function, which we denote exp:
exp(x) = y ⇔ ln y = x.
The cancellation equations for ln and exp are
exp(ln x) = x and ln(exp x) = x.
In particular we have
exp(0) = 1 since ln 1 = 0, and exp(1) = e since ln e = 1.
Geometrically, we get the graph of exp by reflecting the graph of ln about the line y = x.
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1
0
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!3
!2
!1
0
1
2
x
The domain of exp is (∞, ∞), the range of ln. The range of exp is (0, ∞), the domain of
ln.
We would like to think that the function exp x is the same as the familiar function ex .
Now, for a rational number r, the third property of ln gives
ln(er ) = r ln e = r.
From this and the definition of exp, i.e., exp(x) = y ⇔ ln y = x, it follows for rational
r that
exp(r) = er .
We then define ex for all real x including the irrationals, by
ex = exp(x).
Then we have
ex = y ⇔ ln y = x
and
eln x = x for x > 0, and ln(ex ) = x for all x.
The exponential function ex is an positive increasing continuous function with domain
R, range (0, ∞), the limits
lim ex = ∞, and lim ex = 0.
x→∞
x→∞
The latter limit says that y = ex has a horizontal asymptote x = 0.
Theorem: Properties of the Exponential. If x and y are real numbers, and r is
rational, then
ex
(1) ex+y = ex ey , (2) ex−y = y , (3) (ex )r = erx .
e
Proof of (1) [The others are part of your homework.] We use the first property of the
natural logarithm and the cancellation laws:
ln(ex ey ) = ln ex + ln ey = x + y = ln(ex+y ).
Since ln is one-to-one, then we get ex ey = ex+y .
d x
(e ) = ex .
dx
Proof. That ex is differentiable follows because it is the inverse of the differentiable ln x,
and moreover, the derivative of ln x is never zero.
Theorem: Derivative of the Exponential Function.
We need to worry about the derivative not being zero, because if it were at some point
then the inverse function would have an vertical tangent line at the point which means
not differentiable there.
To find the derivative of ex we use implicit differentiation: let y = ex , so then ln y = x.
Since we know that y is a differentiable function of x, we apply the chain rule to ln y = x
to get
d
1 dy
dy
d
(x) =
(ln y) =
⇒
= y = ex .
1=
dx
dx
y dx
dx
Thus we found the derivative.
How many differentiable functions are there whose derivatives are themselves? Just scalar
multiples of ex !
General Logarithm and Exponential Functions. How do we define the power of
a number?
We use the cancellation laws of the exponential and logarithm functions: for a > 0 and
a rational r we define the exponential function of base a to be
ar = exp(ln ar ) = exp(r ln a) = er ln a .
This works even when r is irrational, and so we take
ax = ex ln a for all real x.
Theorem: Laws of Exponents. For x and y any real numbers and positive numbers
a and b, we have
(1) ax+y = ax ay , (2) ax−y =
ax
, (3) (ax )y = axy , (4) (ab)x = ax bx .
ay
Proof of (1) and (3). We convert all of the exponents into base e and apply the laws for
ex :
ax+y = exp (x + y) ln a = exp(x ln a + y ln a) = exp(x ln a) exp(y ln a) = ax ay .
Again using the properties of ex , we have
(ax )y = (exp(x ln a))y = exp(xy ln a) = exp(ln axy ) = axy .
The proof of properties (2) and (4) are for you.
We extend Property (3) of the natural logarithm to any real r:
ln ar = r ln a.
We define the logarithmic function of base a by
loga x = y ⇔ ay = x.
All of the well-known properties of exponentials and logarithms of base a follows from
these definitions.
In fact, the function loga x is just a scalar multiple of ln x: because ay = x, then y ln a =
ln x, and so
ln x
.
loga x = y =
ln a