Post on 09-Apr-2018
MATH 136 The Natural Logarithm Function Definition. The natural logarithm function, denoted by
€
ln x , is defined for all
€
x > 0 by
ln x = 1t1
x∫ dt .
1
y = 1 / t
x
The natural logarithm of
€
x is the area under the graph of y = 1 / t between 1 and
€
x .
Initial Properties
(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 . (iii)
€
ln x < 0 when 0 <
€
x < 1, since the integral from 1 to
€
x will be in reverse direction from right to left.
The Derivative of y =
€
lnx
For ln x = 1t1
x∫ dt , we can apply the 2nd Fundamental Theorem of Calculus to evaluate
the derivative which will simply be the function being integrated re-written as a function of
€
x . Thus,
d ln x( )dx
= d 1
t1
x∫ dt
dx = 1
x, for
€
x > 0 .
Applying the Chain Rule, we have, for g(x ) > 0 , that d ln(g(x)( )dx
= ′ g (x)g(x)
.
Example 1. Let f (x) = ln(x2 + 4) . Then ′ f (x) = 2x
x2 + 4 for all
€
x .
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 xdx
=1x×d xdx
=1x×xx
=1x
, for
€
x ≠ 0.
In other words, when evaluating the derivative, we can ignore the absolute value.
So if y = ln g(x) , then ′ y = ′ g (x)g(x)
, provided g(x ) ≠ 0 .
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
than just
€
x for which
€
cos x > 0 . Then ′ f (x ) =1
cos x×
d(cos x )dx
=− sin xcos x
= − tan x , provided
€
cos x ≠ 0 .
Exponential Property One of the most important properties of logarithms is the identity
ln xr = r ln x
for all
€
x > 0 and for all real numbers r .
To see this result, fix r and let f ( x) = ln xr and g(x ) = r ln x , each for
€
x > 0. Then
f (1) = 0 = g(1) . Moreover, for all
€
x > 0, ′ f (x ) =r xr−1
xr =rx
= ′ g (x ). So f and g have the
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 .
Harmonic Series 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:
1nn=1
∞
∑ = 1 + 12+13+14+15+16+ . . . = +∞
By omitting the first term, we also have 1
2+13+14+15+16+ . . . = +∞ .
Relation to the Natural Logarithm
The natural logarithm function is defined to be the area between the graph of y = 1t
and the
€
t -axis on the interval [1,
€
x].
1 x
y = 1/t
Area = lnx
t-axis
But what happens as
€
x tends to infinity? Is the asymptotic area finite or infinite?
1
Finite or Infinite Area?
1 2 3 4
1/2
1/31/4
We see that the asymptotic area is more than 1
2+13+14+15+16+ . . . , which equals
+∞ by Bernoulli’s argument. Hence, limx→∞
ln x = +∞.
And what about lim
x→0+ln x ? As
€
x→ 0+, then
€
x−1→∞; thus, ln( x−1)→ ∞ and
therefore − ln( x−1 )→ −∞ . We then have
limx→0+
ln x = limx→0+
(−1)(−1) ln x = limx→0+
− ln x−1 = −∞ .
The Graph of y =
€
lnx
The function f (x) = ln x is defined only for
€
x > 0. Also
€
ln1 = 0;
€
ln x > 0 when
€
x > 1;
and
€
ln x < 0 when 0 <
€
x < 1. Since
€
x must be positive, ′ f (x) = 1x
> 0. Thus,
€
ln x is an increasing function and therefore it is also a one-to-one function that has an inverse. Since ′ ′ f (x) =
−1x2 < 0, then
€
ln x must be concave down. Also
€
ln x→∞ as
€
x→∞ and
€
ln x→−∞ as
€
x→ 0+. Below are the graphs of the functions y = ln x and y = ln x .
y = ln x
Domain: (0, ∞) Range: (–∞, ∞)
y = ln x
Domain: x ≠ 0 Range: (–∞, ∞)
The number
€
e and the Function
€
ex 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 . . .
1
y = 1 / t
e
Area = 1
The natural exponential function is defined by y =
€
ex . Because the base
€
e is larger than 1,
€
ex is an exponential growth function. Moreover, we assert that
€
ex is the inverse of
€
ln x .
y = ex
Domain: (–∞, ∞) Range: (0, ∞)
€
ex and
€
ln x , symmetric about the line y = x
First, ln(ex ) = x ln e = x × 1 = x for all
€
x ; thus, the logarithm function “undoes”
€
ex . Moreover, ln(eln x ) = ln x × ln e = ln x , for
€
x > 0 . So because ln is a one-to-one function, we have
€
eln x = x , for
€
x > 0 . Thus,
€
ex “undoes”
€
ln x . So
€
ex and
€
ln x must be inverses.
Inverse Properties
Since
€
ex and
€
ln x are inverse functions, they enjoy a symbiotic relationship.
(i) Domain(
€
ln x ) = (0, ∞) = Range(
€
ex ) (ii) Range(
€
ln x ) = (–∞, ∞) = Domain(
€
ex ) (iii) The graphs of
€
ex and
€
ln x are symmetric about the line y = x . (iv) The composition of one function with the other results in
€
x . That is,
ln(ex ) =
€
x and eln x =
€
x . The last property allows us to solve exponential and logarithmic equations by applying the inverse function.
Solving Equations
We can now solve exponential and logarithmic equations.
To solve
€
ex =
€
a,
apply the natural logarithm giving
€
x =
€
lna.
To solve
€
ln x =
€
a ,
apply the natural exponential giving
€
x =
€
ea . Example 3. Solve the equations: (a) 4 e
−2x = 36. (b) 40 ln(10 − x) = 100. Solution. (a) 4 e
−2x = 36 gives
€
e−2x = 9, then
€
−2x = ln9. Thus,
€
x =
€
−12ln9 =
€
−ln91/2 =
€
−ln3. (b) 40 ln(10 − x) = 100 gives ln(10 − x) = 2.5, then
€
10 − x =
€
e2.5 and
€
x =
€
10 − e2.5.
Sum and Difference of Logarithms
Finally, we can prove the formulas for the sum and difference of natural logarithms:
ln a + ln b = ln(a b) ln a − ln b = ln ab
First, we have
€
elna+lnb = elna × elnb = a × b . Thus, ln(elna+ln b ) = ln(a × b), which
gives ln a + ln b = ln(a b) . Likewise, we have elna−ln b = elna
elnb=ab
. Taking the natural
logarithm then gives ln a − ln b = ln ab .
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.
Example 4. Let f (x) = ln sin2 xx4 9 − x2
and g(x ) =
ln x7 cos5 x
tan3 x. Find ′ f (x) and ′ g ( x) .
Solution. First, we re-write f as follows:
f (x) = ln(sin2 x ) − ln x4 − ln (9 − x2 )1/2( ) =
€
2ln sin x( ) − 4 ln x −12ln(9 − x2) . Then
′ f (x) = 2 cosx
sin x−4x−
−2x2(9 − x2 )
= 2 cot x − 4x+
x(9 − x2 )
.
Similarly, we re-write g as g(x ) = 7 ln x + 5 ln cos x − 3 ln tan x . We can then ignore the absolute values when taking the derivative:
′ g ( x) = 7x+
5(− sin x)cos x
−3sec2 xtan x
= 7x− 5 tan x − 3sec2 x
tan x.