Natural log
Recall that last class we saw that
Z b
bn+1 − an+1
n
x =
n+1
a
if n 6= −1.
What if n = −1?
Question
Let’s define
x
Z
ln(x) =
1
1
dt.
t
1
t
is continuous for t > 0, so it is integrable,
meaning ln(x) exists for x > 0.
Properties The first three properties below should be easy to see, but
the remaining properties require proof.
(1) ln(1) = 0.
(2) x > 1
=⇒
(3) 0 < x < 1
ln(x) > 0.
=⇒
ln(x) < 0.
(4) ln(xy) = ln(x) + ln(y).
(5) ln(xq ) = q ln(x) for rational numbers q.
x
(6) ln
= ln(x) − ln(y).
y
Z xy
1
Proof of (4). We have that ln(xy) =
dt, so
t
1
d
1 d
y
1
(ln(xy)) =
· (xy) =
= .
dx
xy dx
xy x
|
{z
}
by 1st FTOC and chain rule
Math 30G - Prof. Kindred - Lecture 13
Page 1
Thus, ln(x) and ln(xy) have the same derivative, so it follows from a previous result (corollary to mean value thm) that
ln(xy) = ln(x) + C
for some constant C.
If we let x = 1, we see that
ln(y) = ln(1) + C = 0 + C = C
=⇒
ln(xy) = ln(x) + ln(y).
Proof of (5). We have that ln(xq ) =
Z
xq
1
1
dt, so
t
1 d q
d
qxq−1 q
q
(ln(x )) = q · (x ) =
=
q
dx
x
dx
x
x
|
{z
}
by 1st FTOC and chain rule
where we used the power rule for the second step above. We know that
q ln(x) also has the derivative xq . Thus, ln(xq ) and q ln(x) have the same
derivative, so it follows that
ln(xq ) = q ln(x) + C
for some constant C.
If we let x = 1, we see that
0 = ln(1) = q ln(1) + C = q · 0 + C = C
=⇒
ln(xq ) = q ln(x).
Proof of (6). We have
x
= ln(xy −1) = ln(x) + ln(y −1) by property (4)
ln
y
= ln(x) − ln(y) by property (5).
Properties (4), (5), and (6) form the basis of a slide rule, used prior to the
advent of pocket calculators.
Math 30G - Prof. Kindred - Lecture 13
Page 2
Graph of ln(x):
Question
No!
Is ln(x) bounded?
1
ln n
2
= −n ln(2) → −∞ as n → ∞
ln (2n) = n ln(2) → ∞ as n → ∞.
So ln(x) grows very slowly!
ln( |{z}
1080 ) = 80 ln(10) ≈ 184.
# of atoms
in universe
This explains why we use log scales for fast growing quantities. For example,
the Richter scale uses a base-10 log scale, meaning an earthquake that
measures 5.0 on the Richter scale has a shaking amplitude 10 times larger
than one that measures 4.0.
Example We can show that the harmonic series diverges using the integral test. We know that
Z n
n
X
1
1
and
dx = ln(n) have the same behavior as n → ∞.
k
x
1
k=1
Math 30G - Prof. Kindred - Lecture 13
Page 3
Z
Since ln(n) → ∞ as n → ∞, the integral
the harmonic series,
∞
X
1
k=1
k
1
∞
1
dx diverges, which means
x
diverges.
Definition Let e be the number such the ln(e) = 1, i.e., the number such
that the area under 1t is 1.
We explore this value e and the function ex next class...
Math 30G - Prof. Kindred - Lecture 13
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