Download Lecture 2: The Logarithm Function

Lecture 2: The Logarithm Function
2.1 The natural logarithm
Definition The natural logarithm function is the function with value at a real number
x > 0 given by
Z x
1
log(x) =
dt.
1 t
We may also denote log(x) as ln(x).
Some immediate consequences of the definition are:
Z 1
1
dt = 0.
1. log(1) =
1 t
2. If x > 1,
log(x) =
x
Z
1
1
dt > 0.
t
3. If 0 < x < 1,
log(x) =
x
Z
1
1
dt = −
t
Z
1
x
1
dt < 0.
t
4. Elementary area considerations show that
1
3
< log(2) < .
2
4
5. For all x > 0,
d
d
log(x) =
dx
dx
Example
1
1
1
dt = .
t
x
If g(x) = log(3x + 4), then
g 0 (x) =
Example
x
Z
3
.
3x + 4
For x < 0,
1
1
d
log(−x) =
(−1) = .
dx
−x
x
2-1
Lecture 2: The Logarithm Function
2-2
Hence, for any x 6= 0,
d
1
log |x| = .
dx
x
Note: In general, if g(x) = log |f (x)|, then
g 0 (x) =
Example
f 0 (x)
.
f (x)
If f (x) = log(x2 + 1), then
f 0 (x) =
2x
.
+1
x2
2.2 More properties of the logarithm function
Proposition
For any real numbers x > 0 and y > 0,
log(xy) = log(x) + log(y).
Proof
Let f (x) = log(ax), where a > 0 is a constant. Then
f 0 (x) =
a
1
= .
ax
x
Thus the functions f and g(x) = log(x) have the same derivative. Hence there exists a
constant c such that
log(ax) = log(x) + c
for all x > 0. Evaluating at x = 1, we have
log(a) = log(1) + c = c.
Hence
log(ax) = log(a) + log(x).
Proposition
For any real number x > 0,
1
log
= − log(x).
x
Lecture 2: The Logarithm Function
Proof
2-3
We have
1
1
0 = log(1) = log x ×
= log(x) + log
.
x
x
Hence
1
log
= − log(x).
x
Proposition
Proof
For any real numbers x > 0 and y > 0,
x
log
= log(x) − log(y).
y
We have
x
1
log
= log(x) + log
= log(x) − log(y).
y
y
Proposition
If x > 0 and r is a rational number, then
log(xr ) = r log(x).
Proof The proposition clearly holds for r = 0. Suppose r 6= 0. Let f (x) = log(xr ) and
let g(x) = r log(x). Then
f 0 (x) =
r
1
rxr−1 = = g 0 (x).
r
x
x
Hence there exists a constant c such that
log(xr ) = r log(x) + c
for all x > 0. Evaluating at x = 1, we have
0 = r log(1) + c = c.
Hence
log(xr ) = r log(x).
Example
We have
log
x+2
x−3
= log(x + 2) − log(x − 3).
Lecture 2: The Logarithm Function
Hence
d
log
dx
Example
x+2
x−3
=
1
1
−
.
x+2 x−3
We have
log
r
x2 + 1
1
1
= log(x2 + 1) − log(3x + 4).
3x + 4
2
2
Hence
d
log
dx
r
x2 + 1
x
3
= 2
−
.
3x + 4
x + 1 2(3x + 4)
2.3 The graph of the logarithm function
First note that for all x > 0,
d
1
log(x) = > 0,
dx
x
so log(x) is an increasing function on (0, ∞). Since log(2) > 12 , we have
log(2n ) = n log(2) >
n
.
2
Hence
lim log(2n ) = ∞.
n→∞
Since log(x) is an increasing function, it follows that
lim log(x) = ∞.
x→∞
Also,
1
lim log
= − lim log(x) = −∞,
x→∞
x→∞
x
which is equivalent to
lim log(x) = −∞.
x→0+
In particular, log(x) has domain (0, ∞) and range (−∞, ∞). Finally,
d2
1
log(x) = − 2 < 0
2
dx
x
2-4
Lecture 2: The Logarithm Function
2-5
4
3
2
1
10
20
30
40
50
-1
-2
-3
-4
Graph of f (x) = log(x)
for all x > 0, so the graph of log(x) is concave down on (0, ∞). These observations explain
the shape of the graph above.
2.4 Some integrals
Note that the differentiation formula
d
1
log |x| =
dx
x
gives the integration formula
Z
Example
We have
Z
Example
1
dx = log |x| + c.
x
1
1
dx = log |3x + 4| + c.
3x + 4
3
Using the substitution u = x2 + 1, we have
Z
x
1
dx
=
x2 + 1
2
Z
1
1
1
du = log |u| + c = log(x2 + 1) + c.
u
2
2
Lecture 2: The Logarithm Function
Example
2-6
Using the substitution u = cos(x), we have
Z
Z
sin(x)
dx
cos(x)
Z
1
=−
du
u
= − log |u| + c
= − log | cos(x)| + c
tan(x)dx =
= log | sec(x)| + c.
2.5 The number e
Since log(x) is continuous with log(1) = 0 < 1 and log(8) = 3 log(2) > 1, it follows, by the
Intermediate Value Theorem, that there exists a number e such that log(e) = 1. It may
be shown that e is an irrational number with an approximation to 9 decimal places given
by 2.718281828.
In other words, e is the number such that
Z
e
1
Example
1
dx = 1.
x
Using the substitution u = log(x), we have
e
Z
1
log(x)
dx =
x
Z
0
1
udu =
1 2 1
1
u = .
2
2
0