Calculus Of One Real Variable
By Pheng Kim Ving
Chapter 7: The Exponential And Logarithmic Functions
Section 7.2: The Natural Logarithm Function

 

7.2
The Natural Logarithm Function

 

 

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1. The Natural Logarithm Function

 

Consider the relation 23 = 8. We know that 2 is the cube root of 8 and that 8 is the cube of 2. What about 3? Well, in this case,
it's the exponent to which 2 is raised to get 8. It's given the name of “logarithm of base 2 of 8”. Why isn’t it called just the
“exponent to which 2 is raised to get 8”? Of course it can be called just that, because it is just that. Now consider the relation
2x = y. This relation defines y as an exponential function of base 2 of x. Of course x is the exponent to which 2 is raised to get
y. Now consider the inverse function, ie, the function that defines x as the exponent to which 2 is raised to get y: x =
exponent-to-which-2-is-raised-to-get( y) (what a long name of a function!!). If we call that function the “exponent function of
base 2”, there'll be confusion between it and the exponential function of base 2: y = 2x, which is exactly its opposite (inverse).
So we don't call it the “exponent function of base 2”, but rather the “logarithm function of base 2”. Thus in 23 = 8, 3 is given
the name “logarithm of base 2 of 8”.

 

For the definition and discussion of inverse functions, see Section 3.4.

 

 

Fig. 1.1

 

Graph Of y = ln x Is Reflection Of Graph Of Its Inverse y = ex In The Line
y = x.

 

reflection of the graph of its inverse y = ex in the line y = x. It's sketched in Fig. 1.1.

 

If y = ln x, then x = ey. Thus y is the exponent to which e is raised to get x. This means that, since y = ln x, the natural
logarithm of x is the exponent to which e is raised to get x. For simplicity, we'll say “the exponent on e to get x”. Since e is
the base which is raised to y to get x, the natural logarithm is the logarithm of base e.

 

Definition 1.1 - The Natural Logarithm Function

 

The inverse function of the natural exponential function ex is called the natural logarithm function and is denoted ln x, where “ln” is pronounced like the word “lawn”. So:

 

 

So ln x = exponent on e to get x.

 

 

Remark 1.1

 

In “... the exponent on e to get x”, the quantity appearing after “to get”, x in this case, isn't the function (output), it's the
variable (input).

 

Composition

 

Recall that the composition of a function and its inverse is the identity function: f ( f -1(x)) = x and f -1( f (x)) = x. So ln ex = x
and eln x = x. We can see them intuitively as follows. Of course ln ex is the exponent on e to get ex, which is e raised to the
exponent x, thus that exponent must be x. As for eln x, it's e raised to a number that's the exponent on e to get x, consequently
e raised to that exponent must be x. They can be proved formally as follows.

 

For ln ex = x. Let y = ln ex. Then ex = ey, hence x = y, it follows that ln ex = x.
For eln x = x. Let y = eln x. Then ln x = ln y, hence x = y, it follows that eln x = x.

 

Composition

 

ln ex = x,     for all x,
eln x = x,     for all x > 0.

 

 

Properties

 

 

    So in y = ln x, it must be that x > 0. This can also be seen from the fact that x = ey and ey > 0 for all y.

 

b. We have:

 

    ln 1 = (exponent on e to get 1) = 0, since e0 = 1,
    ln e = (exponent on e to get e) = 1, since e1 = e.

 

    We can prove these values as follows. Let a = ln 1. Then ea = 1 = e0. So ln 1 = a = 0, where the last equation is obtained
    from the property that the function ex is one-to-one. Now let b = ln e. Then eb = e = e1. So ln e = b = 1.

 

    Or, using the definition of inverse functions: e0 = 1, thus 0 = ln 1, or ln 1 = 0; now e1 = e, thus 1 = ln e, or ln e = 1. This of
    course is simpler.

 

    These values are also marked in Fig. 1.1, together with the values e0 = 1 and e1 = e.

 

    Now let y = ln x, so that x = ey.

    If 0 < x < 1, then 0 < ey < 1, which requires that y < 0, thus ln x < 0.
    If x > 1, then ey > 1, which requires that y > 0, thus ln x > 0.

 

Properties

 

Refer to Fig. 1.1.

 

 

b. If 0 < x < 1 then ln x < 0,
    ln 1 = 0,
    if x > 1 then ln x > 0,
    ln e = 1.

 

 

The Function L(x)

 

In Section 7.1 Part 2, we defined the function L(x) as the signed area bounded by y = 1/x, the x-axis, and the y-axis. Refer to
Figs. 1.2 and 1.3. The function A(x) is the area of the colored region. The function L(x) was defined as follows:

 

 

We also showed that the inverse of L(x) is ex. So the inverse of ex is L(x). Since the inverse of a function is unique, the natural
logarithm function ln x is the same function as L(x). Now you see why we called that function L(x): we used the letter “L

Fig. 1.2

 

 

 

Fig. 1.3

 

L(x) = –A(x) if 0 < x < 1.

 

because we had the word “Logarithm” in mind.

 

Natural Logarithm As Signed Area

 

Consider the curve y = 1/t for t > 0. See Figs. 1.2 and 1.3. For any point x > 0, let A(x) be the area of the plane region bounded by the curve y = 1/t, the t-axis, the vertical line t = 1, and the vertical line t = x. Let the function L(x) be defined as follows:

 

 

Then L(x) is the natural logarithm function.

 

 

In fact, another approach to introduce the exponential and logarithmic functions in calculus is to present the natural logarithm
first, defined exactly as we did L(x), called natural logarithm instead of L, then present the natural exponential as the inverse
of the natural logarithm, then present the general exponential and logarithm. To show that the general exponential bx is
differentiable and to establish its other properties, either L(x) or the natural logarithm must be introduced before any
exponential, because L(x) or the natural logarithm is needed to show that the number e exists, which in turn leads to the fact
of the differentiability of bx, and to prove its other properties.

 

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2. Properties

 

One of the properties of the natural logarithm that we're going to state and prove is ln xy = ln x + ln y. This relation
expresses the natural logarithm of the product xy of 2 numbers x and y in terms of the natural logarithms of the original
numbers x and y. This property says that the natural logarithm of the product of 2 numbers equals the sum of the natural
logarithms of the original numbers. Intuitively, ln x + ln y is the sum of the exponent on e to get x, say s, and the exponent
on e to get y, say t, so that ln x + ln y = s + t. Now es+t = eset = xy, which indicates that s + t is the exponent on e to get xy.
Thus s + t = ln xy. As a consequence, ln xy = ln x + ln y.

 

Theorem 2.1 – Properties Of Natural Logarithm

 

For any x > 0, any y > 0, and any t we have:

 

 

 

Proof

a. Let s = ln x and t = ln y, so that x = es and y = et. Then ln xy = ln eset = ln es+t = s + t = ln x + ln y.

 

b. ln (1/x) + ln x = ln ((1/x)x) = ln 1 = 0, where the first equation is obtained by part c. So ln (1/x) = - ln x.

c.
ln (x/y) = ln (x(1/y)) = ln x + ln (1/y) = ln x - ln y, using parts c and d.

 

d. Note that if t is a positive integer then, by repetitive applications of part c:

 

   

 

    For t = 0, we have ln x0 = ln 1 = 0 = 0 ln x.

 

    For any non-0 real number t. Let y = ln xt, so that xt = ey. Then x = ey/t, then y/t = ln x, then y = t ln x. Thus ln xt =
   
t ln x.

 

    Consequently, for any real number t we have ln xt = t ln x.

EOP

 

Corollary 2.1 - Natural Logarithms Of Extended Products And Quotients

 

For any real pi, any real qj, any positive integer m, and any positive integer n we have:

 

 

 

Proof

EOP

 

Example 2.1

 

Simplify the following expressions.
a.  ln 4 + 3 ln 2 – 2 ln 8.
b.  ln(x + 1)2ln(x3 + x2 – 3x – 3) + ln(x2 – 3).

 

Solution

EOS

 

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3. Differentiation

 

The derivative of L(x) is 1/x. So the derivative of ln x is 1/x. Of course the proof for the case of ln x uses the fact that ln x is
the inverse of ex. It's much shorter and much simpler than the proof for the case of L(x) as carried out in Section 7.1 Part 2.

 

Theorem 3.1 - Derivative Of Natural Logarithm

 

For all x > 0 we have:

 

 

 

Proof
Let
y = ln x. Then x = ey. Differentiating this equation implicitly with respect to x we have:

 

EOP

Derivative Of ln |x|

 

 

Corollary 3.1 - Derivative Of Natural Logarithm Of Absolute Value

 

 

 

Proof
It remains to prove that (d/dx) ln |x| = 1/x if x < 0. For any x < 0, using the chain rule we have:

 

EOP

 

Example 3.1

 

Find the following derivatives. Re-state your answers in the form of indefinite integrals.

 

 

Solution

EOS

 

Remark 3.1

 

 

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4. Graph

 

y = ln x.

 

Inflection Points. Since there's no change of concavity, there's no inflection point.
Special Characteristic. ln e = 1.

 

The graph of y = ln x is sketched in Fig. 4.1.

 

Fig. 4.1

 

Graph Of y = ln x.

 

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5. Slowness

 

From Graph

 

Refer to Fig. 5.1. When x increases from the near right of 0 to 1, ln x increases as much as from near minus infinity to 1. This
corresponds to the fact that when x increases as much as from near minus infinity to 0, ex increases only from the near above

Fig. 5.1

 

The natural exponential y = ex increases toward infinity very fast, so its
inverse the natural logarithm y = ln x increases toward infinity very slowly.

 

of 0 to 1. But when x is greater than 1 and increases, ln x increases very slowly. This corresponds to the fact that when x is
positive and increases, ex increases very fast. In fact, ln x increases more and more slowly, as evidenced by the fact that its
graph becomes less and less steep.

 

From Table Of Values

 

Now let's reach for a calculator and calculate the exponents on e as displayed in the table in Fig. 5.2. Clearly as x increases
from 1 to 1,000,000, ln x increases only from 0 to 13.82. Very slow indeed. Of course in the table we can find the exponents
on e by trial and error. However it's faster to just calculate the value of ln x to use as the exponent on e to get x, as we
actually did for this table!! Of course ln x is the exponent on e to get x.

 

Fig. 5.2

 

As x increases from 1 to 1,000,000, ln x increases
only from 0 to 13.82.

 

From Inverse

 

Suppose y = f (x). If x changing by 1 unit implies y changing by 3 units, then y changing by 1 unit implies x changing by 1/3
unit. If y changes faster than x does, then of course x changes slower than y does. Now the inverse function is x = f -1( y).
Here again if y changes faster than x does, then of course x changes slower than y does. Conforming to the good old tradition,
we now use the letter x for the variable and the letter y for the function (for more detail see Section 3.4). So now for the
inverse function we have y = f -1(x). Thus if the y in y = f (x) changes faster than the x there does, then the y in y = f -1(x)
changes slower than the x there does. In brief, if a function changes fast, then its inverse changes slowly.

 

We saw in Section 7.1 Part 9 that the natural exponential function ex grows very fast toward infinity, as asserted by this limit:

 

 

Consequently, its inverse the natural logarithm function must be very slow in its increase toward infinity.

 

Theorem 5.1 - Slowness Of Growth Of Natural Logarithm

 

 a.  ln x < x     for all x > 0. (In Fig. 5.1, the entire graph of y = ln x is below that of y = x.)

 

 

 

Proof

EOP

 

Remarks 5.1 (On Theorem 5.1)

 

a. For all x > 0, ln x grows slower than x does.

 

b. By part a, ln xt grows slower than xt does. If 0 < t < 1, then xt/t > xt, so ln xt must grow slower than xt/t does. If t > 1, then
    xt/t < xt, so we see that not only does ln xt grow slower than xt does, but even slower than xt/t, which is smaller than xt,
    does.

 

 

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Problems & Solutions

 

1.  Simplify the following expressions.

 

a.  4 ln 2 + ln 3 – ln 6.
b.  ln(x + 1) + ln(x – 1) – ln(x2 – 1).
c.  ln(1 + cos x) + ln(1 – cos x).

 

Solution

 

a.  4 ln 2 + ln 3 – ln 6 = ln 24 + ln(3/6) = ln 16 + ln(1/2) = ln(16 x 1/2) = ln 8.

 

b.  ln(x + 1) + ln(x – 1) – ln(x2 – 1) = ln((x + 1)(x – 1)/(x2 – 1)) = ln((x2 – 1)/(x2 – 1)) = ln 1 = 0.

 

c.  ln(1 + cos x) + ln(1 – cos x) = ln((1 + cos x)(1 – cos x)) = ln(1 – cos2x) = ln sin2 x = 2 ln sin x.

 

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2.  Differentiate the following functions, simplify your answers whenever possible, and re-state the results as indefinite
     integrals.

 

a.  y = ln((2x + 3)/(x2 + 4)), x > –3/2, so that (2x + 3)/(x2 + 4) > 0.
b.  y = ln |csc x + cot x|.
c.
  y = ln2 ln x (ie, ( ln ln x)2).

 

Solution

 

 

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3.  Find:

 

 

Solution

 

a.  f '(x) = A cos ln x + Ax (– sin ln x)(1/x) + B sin ln x + Bx (cos ln x)(1/x) = (A + B ) cos ln x + (BA) sin ln x.

 

b.  Part a shows that if B = A, then we would get rid of sin ln x. So let g(x) = Ax cos ln x + Ax sin ln x. Then g'(x) =
     2A cos ln x. It follows that:

 

 

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Solution

 

 

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5.  Prove that for any t > 0:

 

   

 

Note

 

 

can be written as:

 

 

Solution

 

 

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