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.