MATH 136
The Natural Logarithm Function
Definition. The natural logarithm function, denoted by ln x , is defined for all x > 0 by
x 1
ln x = ∫ dt .
1 t
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y =1/ t
1
x
The natural logarithm of x is the area
under the graph of y = 1 / t between 1 and x .
Initial Properties
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(i) ln1 = 0, since the area of the line segment from 1 to 1€has no area.
(ii) ln x > 0 when x > 1, since there will be positive area from 1 to x .
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(iii) ln x < 0 when 0 < x < 1, since the integral from 1 to x will be in reverse direction
from right to left.
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The Derivative of y = ln x
€
x 1€
For ln x = ∫ dt , we can apply the 2nd Fundamental Theorem of Calculus to evaluate
1 t
€
the derivative which will simply be the function
being integrated re-written as a
function of x . Thus,
x 1 
d  ∫ dt 
1
1 t 
d (ln x )
=
= , for x > 0 .
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x
dx
dx
Applying the Chain Rule, we have, for g(x ) > 0 , that
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Example 1. Let f (x) = ln(x 2 + 4) . Then f ′(x) =
d (ln(g(x)) g ′(x)
=
.
g(x)
dx
2x
for all x .
x +4
2
The Derivative of y =€ln x
If f ( x) = ln x , then f is defined for all x ≠ 0 . By the Chain Rule, f ′ (x ) is given by
d ln x
x
1 d x
1
1
=
×
=
×
= , for x ≠ 0 .
dx € x
dx
x
x
x
€
€
In other words, when evaluating the derivative, we can ignore the absolute value.
g ′(x)
So if y = ln g(x) , then y ′ =
, provided g(x ) ≠ 0 .
g(x)
We often use the absolute value in order to expand the domain of the function.
Example 2. Let f ( x) = ln cos x . Then f is defined for all x for which cos x ≠ 0 , rather
1
d(cos x ) − sin x
than just x for which cos x > 0 .
Then f ′ (x ) =
×
=
= − tan x ,
cos x
dx
cos x
provided cos x ≠ 0 .
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Exponential Property
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€ of the most important properties of logarithms is the identity
One
ln x r = r ln x
for all x > 0 and for all real numbers r .
To see this result, €fix r and let f ( x) = ln x r and g(x ) = r ln x , each for x > 0 . Then
r x r −1 r
= = g′ (x ) . So f and g have the
f (1) = 0 = g(1) . Moreover, for all x > 0 , f ′ (x ) =
x
xr
same value at x = 1, and then grow at the same rate due to equal
€ derivatives. These
facts are enough to force f ( x) = g( x) for all x > 0 .
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Harmonic Series
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In 1689, Johann Bernoulli gave a proof of the divergence of the harmonic series. That is,
he proved that the sum of the reciprocals of the natural numbers was infinite:
∞
1
1 1 1 1 1
= 1 + + + + + + . . . = +∞
2 3 4 5 6
n =1 n
∑
By omitting the first term, we also have
1 1 1 1 1
+ + + + + . . . = +∞.
2 3 4 5 6
Relation to the Natural Logarithm
The natural logarithm function is defined to be the area between the graph of y =
the t -axis on the interval [1, x ].
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1
and
t
Area = lnx
y = 1/t
1
t-axis
x
But what happens as x tends to infinity? Is the asymptotic area finite or infinite?
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Finite or Infinite Area?
1
1/2
1/3
1/4
1
2
3
4
1 1 1 1 1
+ + + + + . . . , which equals
2 3 4 5 6
+∞ by Bernoulli’s argument. Hence, lim ln x = +∞.
We see that the asymptotic area is more than
x →∞
And what about
lim ln x ?
x →0+
As x → 0+ , then x −1 → ∞; thus, ln( x −1 ) → ∞ and
therefore − ln( x −1 ) → −∞ . We then have
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lim ln x = lim (−1)(−1) ln x = lim − ln x −1 = −∞ .
x →0+
x →0+
x →0+
The Graph of y = ln x
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The function f (x) = ln x is defined only for x > 0. Also ln1 = 0; ln x > 0 when x > 1;
1
and ln x < 0 when 0 < x < 1. Since x must
> 0. Thus, ln x is an
€ be positive, f ′(x) =
x
increasing function and therefore it is also a one-to-one function that has an inverse.
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−1
Since f ′′(x) = 2 < 0, then ln x must be concave down. Also ln x → ∞ as x → ∞ and
€x
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ln x → −∞ as x → 0+ . Below are the graphs of the functions y = ln x and y = ln x .
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y = ln x
Domain: (0, ∞) Range: (–∞, ∞)
y = ln x
Domain: x ≠ 0 Range: (–∞, ∞)
The number e and the Function e x
We define e to be the number such that lne = 1. That is, the area under the graph of
y = 1 / t from 1 to e equals 1. The numerical value of e is approximately 2.718281828 . . .
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y =1/ t
€
Area
=1
1
e
The natural exponential function is defined by y = e x . Because the base e is larger
than 1, e x is an exponential growth function. Moreover, we assert that e x is the inverse
of ln x .
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x
e x and ln x ,
symmetric about the line y = x
y=e
Domain: (–∞, ∞) Range: (0, ∞)
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First, ln(e x ) = x ln e = x × 1 = x for all x ; thus, the logarithm function “undoes” e x .
Moreover, ln(e ln x ) = ln x × ln e = ln x , for x > 0 . So because ln is a one-to-one function,
we have e ln x = x , for x > 0 . Thus, e x “undoes” ln x . So e x and ln x must be inverses.
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Inverse Properties
Since e x and ln x are inverse functions, they enjoy a symbiotic relationship.
(i) Domain( ln x ) = (0, ∞) = Range( e x )
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(ii) Range( ln x ) = (–∞, ∞) = Domain( e x )
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(iii) The graphs of e x and ln x are symmetric about the line y = x .
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(iv) The composition of one function with the other results in x . That is,
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e ln x = x .
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The last property allows us to solve exponential and logarithmic equations by
applying the inverse function.
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ln(e x ) = x
and
Solving Equations
We can now solve exponential and logarithmic equations.
To solve e x = a ,
apply the natural logarithm giving
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x€= ln a .
To solve ln x = a ,
€
apply the natural exponential giving
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x = ea .
−2x
Example 3. Solve the equations:
= 36.
€ (a)€ 4 e
(b) 40 ln(10 − x) = 100.
Solution. (a) 4 e −2x = 36 gives e−2x = 9, then −2x = ln9. Thus,
€
1
x = − ln9 = −ln91/ 2 = −ln 3 .
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(b) 40 ln(10 − x) = 100 gives ln(10 − x) = 2.5, then 10 − x = e 2.5 and x = 10 − e 2.5 .
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Sum and Difference of Logarithms
Finally, we can prove the formulas for the sum and difference of natural logarithms:
a
ln a − ln b = ln  
b
ln a + ln b = ln(a b)
Thus, ln(e ln a +ln b ) = ln(a × b) , which
lna
a
lna−ln b e
= lnb = . Taking the natural
gives ln a + ln b = ln(a b) . Likewise, we have e
b
e
€ gives ln a − ln b = ln  a  .
logarithm then
 b
First, we have e ln a+lnb = e ln a × e lnb = a × b .
Further Derivatives
We now can evaluate derivatives involving the composition of the logarithm with more
complicated functions. But before evaluating a derivative, it is often helpful to re-write
the function as separate logarithms using the arithmetic properties.
 sin 2 x 
x 7 cos 5 x
Example 4. Let f (x) = ln 
and g(x ) = ln
. Find f ′(x) and g ′( x) .

tan 3 x
 x4 9 − x2 
Solution. First, we re-write f as follows:
2
4
(
2 1/2
f (x) = ln(sin x ) − ln x − ln (9 − x )
f ′(x) =
) = 2ln( sin x ) − 4 ln x
1
− ln(9 − x 2 ) . Then
2
2 cosx 4
−2x
4
x
− −
= 2 cot x − +
.
2
sin x
x 2(9 − x )
x (9 − x 2 )
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Similarly, we re-write g as g(x ) = 7 ln x + 5 ln cos x − 3 ln tan x .
ignore the absolute values when taking the derivative:
g ′( x) =
7 5(− sin x) 3sec 2 x 7
3sec 2 x
= − 5 tan x −
.
+
−
x
cos x
tan x
x
tan x
We can then