7.1 The Natural Logarithm Function
| Calculus Of One Real Variable – By Pheng Kim Ving |
| 7.1 |
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The Exponential And Logarithmic Functions
Consider the relation 32 = 9. We see that:
3 = 2nd root (or square root) of 9,
and that:
9 = 2nd power (or square) of 3.
(We use “ 2nd ” to emphasize that 2 is present in both relations between 3 and 9.) What about 2? Well, we see that:
2 = exponent on 3 (exponent to which 3 is raised) to get 9.
We say that 2 is the logarithm of 9 with base 3, and write 2 = log 3 9. The notation log 3 9 is read “ the logarithm of 9 with
base 3”, or, for short, “ log base 3 of 9”, which means exponent on 3 to get 9.
Consider the function y = f(x) = ax. Then x = f –1( y), and also x = (exponent on a to get y) = loga y. This shows that f –1( y) =
loga y. The logarithmic function loga x is the inverse of the exponential function ax.
Keep in mind that the exponential function y = ax is different from the power function y = xa. In xa, the variable is in the
base and the constant is in the exponent, while in ax, the variable is in the exponent and the constant is in the base. Note
that ax is called an exponential function because the variable is in the exponent. Remark that the variable x is the
exponent and the value ax is the exponential.
Let y = ax. Then x = (exponent on a to get y) = loga y. The logarithmic function g(x) = loga x is the inverse of the
exponential function f(x) = ax; see Section 3.4 Part 2 for inverse functions. The distinctions outlined in Fig. 1.1
should be clear.
| Fig. 1.1 Distinctions Between Functions. |
The Exponential Before The Logarithm
In algebra, the exponential function is introduced before the logarithmic function, which is defined as its inverse. That’s
because the value of the exponential function involves the raising of a quantity to a power, a notion that’s already familiar.
Differentiation Of The Exponential Function
Let’s differentiate the exponential function ax. Observe that so far we don’t yet know what (d/dx) ax is. And we cannot
claim that it’s xax–1! Remember, ax is not a power function. We must go back to the definition of the derivative. Let f(x) =
ax. We have:
To estimate the value of e, we give h a small value fairly close to 0, say h = 0.001, and find a value of a that makes the
ratio (ah – 1)/h close to 1. We reach for a calculator and proceed as shown in Fig. 1.2.
| Fig. 1.2 Finding value of a that makes ratio (a0.001 – 1)/0.001 close to 1. |
The Logarithm ln x Before The Exponential ex
In the differentiation of ax as shown above, we assume that ax is differentiable at x = 0. We wish to get rid of the dependency on this assumption in developing the theory of the exponential and logarithmic functions. There are several methods to achieve this purpose. One method, which we’ll use, is to define the logarithmic function first, independently of the exponential function. In introductory calculus, or at least in many introductory calculus texts, the logarithmic function
ln x is introduced before the exponential function ex. The precise sequence that we adopt is as follows.
– ln x, called natural logarithm;
– ex, called natural exponential ; inverse of natural logarithm ln x will be recognized as this natural exponential; so
natural logarithm ln x will be recognized as logarithm with base e;
– ax, any a > 0, called general exponential, defined as generalization of ex;
– loga x, 0 < a < 1 or 1 < a, called general logarithm, defined as inverse of general exponential.
The natural logarithm is presented in this section. The remaining functions are presented in the next 2 sections.
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| 2. The Natural Logarithm Function |
We’ve seen above that (d/dx) ex = ex, assuming that for any a > 0 (including e), ax is differentiable at x = 0. The inverse
of ex is ln x. Let y = ln x, so that x = ey. By Section 3.4 Eq. [5.2] we have (d/dx) ln x = dy/dx = 1/(dx/dy) = 1/ey =
1/x. Wow, if we can find, in a way that’s independent of ax (including ex), a function that turns out to be an antiderivative
of 1/x, then that function would be defined as ln x, and we would get rid of the assumption about ax! But, wait a minute,
does 1/x have an antiderivative that’s a function that we so far already know? Note that 1/x = x–1. Is there an integer
power function that’s an antiderivative of 1/x? To find out, let’s check the integer power functions and their derivatives, as
done in Fig. 2.1. In the function f(x) column, every integer power is present somewhere. However in the derivative f ‘(x)
| Fig. 2.1 We so far don’t yet have an antiderivative of 1/x = x–1. |
column, the power x–1 = 1/x is “annihilated” by 0, and thus is absent. Now we accept that so far we don’t yet have an
antiderivative of 1/x.
So let’s consider the function y = 1/x for x > 0, whose graph is shown in Fig. 2.2. The reason why we use x > 0 only
appears below. For any u > 0, let A be the area of the plane region below the graph of y = 1/x, above the x-axis, and
between the vertical lines x = 1 and x = u. The region is colored in Fig. 2.2. Obviously, as u varies (while remaining
> 0), A also varies. That is, A is a function of u. Thus it’s only natural to denote it A(u). Now we want to denote the
variable of the newly defined function A by the letter x, instead of u, to conform to tradition. In Fig. 2.2 we replace the
| Fig. 2.2 Colored area A(u) is a function of u. |
| Fig. 2.3 |
| Fig. 2.4 ln x = –A(x) if 0 < x < 1. |
letter x by t and the letter u by x, to get the graph of the function y = 1/t and the function A(x), as displayed in Fig. 2.3.
Note that the horizontal axis is re-named as the t-axis, and that instead of the letter t, we can use any other letter that’s
normally used for variables, like u or v, as long as it’s not the already-used x or y.
< 1, and f(1) = 0. When we differentiate f(x), as we do in Theorem 3.1 below, we get (d/dx) f(x) = 1/x. Eureka! We’ve
found our function: it’s that f ! Hence let’s define it as the natural logarithm.
Definition 2.1
| For 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. See Figs. 2.3 and 2.4. The natural logarithm function, denoted ln, which is pronounced as the word |
Clearly this definition of the natural logarithm function is completely independent of any exponential function. In the next
section, we’ll learn that the inverse of the function defined in this definition is the exponential function ex. That’s the
ultimate justification of the defining of this function as the natural logarithm, which is the logarithm with base e.
Reason Why We Use y = 1/x For x > 0 Only. Let y = ln x. Then x = ey. Now ey > 0 for all y. And that’s the reason.
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| 3. Differentiation Of The Natural Logarithm |
Theorem 3.1
| For all x > 0 we have: |
Proof
Let x > 0 be arbitrary. We utilize the definition of the derivative:
| Fig. 3.1 (d/dx) ln x = 1/x. |
By the squeeze theorem {3.1} applied to one-sided limits we get:
EOP
{3.1} See Section 1.1.2 Theorem 5.1.
Remark that in this proof there’s no assumption about the differentiability of ax.
Corollary 3.1
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
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| 4. Properties Of The Natural Logarithm |
Meaning Of xr Where r Is Rational
Property v below deals with xr where r is rational. Let’s define what xr means. We know what it means if r is an integer.
If r is a rational number, say r = m/n where m and n are integers and n > 0, we define xr as follows:
The definition of xr where r is any real number (irrational included) is given in Section 7.3 Definition 1.1 in the form of
the definition of ax where a > 0 and x is any real number.
Property v below states that ln xr = r ln x for any rational number r. As we’ll see in Section 7.3 Eq. [2.6], this identity
also holds true for any real number r, rational or irrational.
Theorem 4.1 – Properties Of The Natural Logarithm
Proof
i. These properties are obvious from the definition of ln x {4.1}.
ii. Let a be an arbitrary constant. We have (d/dx) ( ln ax – ln x) = (d/dx) ln ax – (d/dx) ln x = (1/(ax))a – 1/x = 0. So
ln ax – ln x = C for all x > 0, where C is a constant {4.2}. In particular, at x = 1 we have ln(a . 1) – ln 1 = C, or ln a –
0 = C, yielding C = ln a. Thus, ln ax – ln x = ln a, or ln ax = ln a + ln x. Letting y = a we get ln yx = ln y + ln x, or
ln xy = ln x + ln y.
iii. (d/dx) (ln 1/x + ln x) = (1/(1/x))(–1/x2) + 1/x = 0. Consequently, as in part ii, ln 1/x + ln x = C. At x = 1, we have
ln 1/1 + ln 1 = C, so that C = 0. This leads to ln 1/x = – ln x.
iv. ln x/y = ln x(1/y) = ln x + ln 1/y = ln x – ln y, by parts ii and iii.
v. (d/dx) ( ln xr – r ln x) = (1/xr)(rxr–1) – r/x = 0. Hence, as in part ii, ln xr – r ln x = C. At x = 1 we have ln 1 – r ln 1
= C, so that C = 0. This yields ln xr = r ln x.
{4.1} See Definition 2.1.
{4.2} See Section 5.1 Theorem 6.1.
{4.3} See Section 5.3 Theorem 2.1.
Example 4.1
Simplify the following expressions.
a. ln 4 + 3 ln 2 – 2 ln 8.
b. ln(x + 1)2 – ln(x3 + x2 – 3x – 3) + ln(x2 – 3).
Solution
EOS
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| 5. Graph Of The Natural Logarithm Function |
Let y = ln x.
x-Intercepts. ln x = 0 iff x = 1, so there’s only one x-intercept; it’s x = 1.
y-Intercept. None, because all x is positive.
Limits. See Theorem 4.1 vii above.
| Fig. 5.1 Chart For y = ln x. |
| Fig. 5.2 Graph Of y = ln x. |
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1. 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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2. 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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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 + (B – A) 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:
a. ln x < x for all x > 0.
Solution
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