<<

. 5
( 11)



>>

3’
3’
’2 ’1
(x ’ 3) dx = lim ’(x ’ 3)
lim
’0+ 2 ’0+ 2
’1
’ 1’1 .
= lim
’0+
This last limit is +∞. We therefore conclude that the improper integral diverges.

’1
+ 1)4/5 ) dx.
You Try It: Evaluate the improper integral ’2 (dx/(x
Improper integrals with integrand which is in¬nite at the left endpoint of
integration are handled in a manner similar to the right endpoint case:
EXAMPLE 5.16
Evaluate the integral
1/2
1
dx.
x · ln2 x
0

SOLUTION
This integral is improper with in¬nite integrand at 0. The value of the integral
is de¬ned to be
1/2 1
lim dx,
x · ln x 2
’0+
provided that this limit exists.
Since 1/(x ln2 x) is continuous on the interval [ , 1/2] for > 0, this last
integral can be evaluated directly and will have a ¬nite real value. For clarity,
write •(x) = ln x, • (x) = 1/x. Then the (inde¬nite) integral becomes
• (x)
dx.
• 2 (x)
Clearly the antiderivative is ’1/•(x). Thus we see that
1/2
1/2 1 1 1 1
dx = lim ’ = lim ’ ’’
lim .
x ·ln2 x
’0+ ’0+ ln x ’0+ ln(1/2) ln
136 CHAPTER 5 Indeterminate Forms

Now as ’ 0+ we have ln ’ ’∞ hence 1/ ln ’ 0. We conclude that
the improper integral converges to 1/ ln 2.

+ 2)’1/2 dx.
0
You Try It: Evaluate the improper integral ’2 1/(x
Many times the integrand has a singularity in the middle of the interval of inte-
gration. In these circumstances we divide the integral into two pieces for each of
which the integrand is in¬nite at one endpoint, and evaluate each piece separately.
EXAMPLE 5.17
Evaluate the improper integral
4
4(x + 1)’1/5 dx.
’4

SOLUTION
The integrand is unbounded as x tends to ’1. Therefore we evaluate
separately the two improper integrals
’1 4
’1/5
4(x + 1)’1/5 dx.
4(x + 1) dx and
’4 ’1
The ¬rst of these has the value
’1’
’1’
’1/5
4(x + 1) dx = lim 5(x + 1) 4/5
lim
’0+ ’4 ’0+ ’4
= lim 5 (’ )4/5 ’ (’3)4/5
’0+
= ’5 · 34/5
The second integral has the value
4
4
’1/5
4(x + 1) dx = lim 5(x + 1) 4/5
lim
’0+ ’1+ ’0+ ’1+
= lim 5 54/5 ’ 4/5
’0+
=5 9/5
.
We conclude that the original integral converges and
’1
4 4
’1/5 ’1/5
4(x + 1)’1/5 dx
4(x + 1) dx = 4(x + 1) dx +
’4 ’4 ’1
= ’5 · 3 +5
4/5 9/5
.
’1 dx.
3
You Try It: Evaluate the improper integral ’4 x
137
CHAPTER 5 Indeterminate Forms

It is dangerous to try to save work by not dividing the integral at the singularity.
The next example illustrates what can go wrong.
EXAMPLE 5.18
Evaluate the improper integral
2
x ’4 dx.
’2
SOLUTION
What we should do is divide this problem into the two integrals
0 2
’4
x ’4 dx.
dx and (—)
x
’2 0
Suppose that instead we try to save work and just antidifferentiate:
2
2 1 1
’4
dx = ’ x ’3 =’
x .
3 12
’2 ’2

A glance at Fig. 5.4 shows that something is wrong. The function x ’4 is
positive hence its integral should be positive too. However, since we used
an incorrect method, we got a negative answer.




Fig. 5.4
In fact each of the integrals in line (—) diverges, so by de¬nition the improper
integral
2
x ’4 dx
’2
diverges.
138 CHAPTER 5 Indeterminate Forms

EXAMPLE 5.19
Analyze the integral
1
1
dx.
x(1 ’ x)1/2
0

SOLUTION
The key idea is that we can only handle one singularity at a time. This
integrand is singular at both endpoints 0 and 1. Therefore we divide the domain
of integration somewhere in the middle”at 1/2 say (it does not really matter
where we divide)”and treat the two singularities separately.
First we treat the integral
1/2 1
dx.
x(1 ’ x)1/2
0
Since the integrand has a singularity at 0, we consider
1/2



Y
1
lim dx.
x(1 ’ x)1/2
’0+
FL
This is a tricky integral to evaluate directly. But notice that
1 1
AM


x(1 ’ x)1/2 x · (1)1/2
¤ x ¤ 1/2. Thus
when 0 <
TE



1/2 1/2 1/2
1 1 1
dx ≥ dx = dx.
x(1 ’ x)1/2 x · (1)1/2 x
We evaluate the integral: it equals ln(1/2) ’ ln . Finally,
lim ’ ln = +∞.
’0+
The ¬rst of our integrals therefore diverges.
But the full integral
1 1
dx
x(1 ’ x)1/2
0
converges if and only if each of the component integrals
1/2 1
dx
x(1 ’ x)1/2
0
and
1 1
dx
x(1 ’ x)1/2
1/2
139
CHAPTER 5 Indeterminate Forms

converges. Since the ¬rst integral diverges, we conclude that the original
integral diverges as well.

’1/3 dx
3
You Try It: Calculate as an improper integral.
’2 (2x)


5.3.3 AN APPLICATION TO AREA
Suppose that f is a non-negative, continuous function on the interval (a, b] which
is unbounded as x ’ a + . Look at Fig. 5.5. Let us consider the area under the graph
of f and above the x-axis over the interval (a, b]. The area of the part of the region
over the interval [a + , b], > 0, is
b
f (x) dx.
a+




Fig. 5.5

Therefore it is natural to consider the area of the entire region, over the interval
(a, b], to be
b
lim f (x) dx.
’0+ a+

This is just the improper integral
b
Area = f (x) dx.
a
140 CHAPTER 5 Indeterminate Forms

EXAMPLE 5.20
Calculate the area above the x-axis and under the curve
1
y= 0 < x ¤ 1/2.
,
x · ln4/3 x

SOLUTION
According to the preceding discussion, this area is equal to the value of the
improper integral
1/2 1/2
1 1
dx = lim dx.
x · ln4/3 x x · ln4/3 x
’0+
0

For clarity we let •(x) = ln x, • (x) = 1/x. Then the (inde¬nite) integral
becomes
3
• (x)
dx = ’ 1/3 .
• 4/3 (x) • (x)

Thus
1/2
1/2 1 3
dx = lim ’
lim
x · ln4/3 x ln1/3 x
’0+ ’0+

’3 ’3
= lim ’ .
[’ ln 2]1/3 [ln ]1/3
’0+

Now as ’ 0 then ln ’ ’∞ hence 1/[ln ]1/3 ’ 0. We conclude that our
improper integral converges and the area under the curve and above the x-axis
equals 3/[ln 2]1/3 .




5.4 More on Improper Integrals
5.4.1 INTRODUCTION
Suppose that we want to calculate the integral of a continuous function f (x) over
an unbounded interval of the form [A, +∞) or (’∞, B]. The theory of the integral
that we learned earlier does not cover this situation, and some new concepts are
needed. We treat improper integrals on in¬nite intervals in this section, and give
some applications at the end.
141
CHAPTER 5 Indeterminate Forms

5.4.2 THE INTEGRAL ON AN INFINITE INTERVAL
Let f be a continuous function whose domain contains an interval of the form
[A, +∞). The value of the improper integral
+∞
f (x) dx
A
is de¬ned to be
N
lim f (x) dx.
N ’+∞ A

Similarly, we have: Let g be a continuous function whose domain contains an
interval of the form (’∞, B]. The value of the improper integral
B
g(x) dx
’∞
is de¬ned to be
B
lim f (x) dx.
M’’∞ M

EXAMPLE 5.21
Calculate the improper integral
+∞
x ’3 dx.
1

SOLUTION
We do this problem by evaluating the limit
N N
x ’3 dx = ’(1/2)x ’2
lim lim
N ’+∞ 1 1
N’+∞

lim ’(1/2) N ’2 ’ 1’2
=
N’+∞
1
=.
2
We conclude that the integral converges and has value 1/2.
EXAMPLE 5.22
Evaluate the improper integral
’32
x ’1/5 dx.
’∞
142 CHAPTER 5 Indeterminate Forms

SOLUTION
We do this problem by evaluating the limit

5 4/5 ’32
’32
’1/5
dx = lim
lim x x
M’’∞ 4
M’’∞ M M
5
= lim (’32)4/5 ’ M 4/5
M’’∞ 4
5
= lim 16 ’ M 4/5 .
M’’∞ 4

This limit equals ’∞. Therefore the integral diverges.
∞ ’3 dx.
1 (1 + x)
You Try It: Evaluate
Sometimes we have occasion to evaluate a doubly in¬nite integral. We do so by
breaking the integral up into two separate improper integrals, each of which can be
evaluated with just one limit.
EXAMPLE 5.23
Evaluate the improper integral
∞ 1
dx.
’∞ 1 + x
2


SOLUTION
The interval of integration is (’∞, +∞). To evaluate this integral, we break
the interval up into two pieces:
(’∞, +∞) = (’∞, 0] ∪ [0, +∞).
(The choice of zero as a place to break the interval is not important; any
other point would do in this example.) Thus we will evaluate separately the
integrals
+∞ 0
1 1
dx and dx.
1 + x2 ’∞ 1 + x
2
0
For the ¬rst one we consider the limit
N 1 N
dx = lim Tan’1 x
lim
1+x 2
N ’+∞ 0 N ’+∞ 0

Tan’1 N ’ Tan’1 0
= lim
N ’+∞
π
= .
2
143
CHAPTER 5 Indeterminate Forms

The second integral is evaluated similarly:
0 1 0
dx = lim Tan’1 x
lim
1+x 2
M’’∞ M M’’∞ M

Tan’1 0 ’ Tan’1 M
= lim
M’’∞
π
= .
2
Since each of the integrals on the half line is convergent, we conclude that the
original improper integral over the entire real line is convergent and that its
value is
π π
+ = π.
2 2
∞ ’1 dx.
1 (1 + x)
You Try It: Discuss

5.4.3 SOME APPLICATIONS
Now we may use improper integrals over in¬nite intervals to calculate area.
EXAMPLE 5.24
Calculate the area under the curve y = 1/[x · (ln x)4 ] and above the x-axis,
2 ¤ x < ∞.
SOLUTION
The area is given by the improper integral
+∞ N
1 1
dx = lim dx.
x · (ln x)4 x · (ln x)4
N ’+∞
2 2
For clarity, we let •(x) = ln x, • (x) = 1/x. Thus the (inde¬nite) integral
becomes
1/3
• (x)
dx = ’ 3 .
• 4 (x) • (x)
Thus
1/3 N
N 1
dx = lim ’ 3
lim
x · (ln x)4
N ’+∞ 2 ln x 2
N’+∞
1/3 1/3
= lim ’ ’3
ln3 N ln 2
N’+∞
1/3
= 3.
ln 2
Thus the area under the curve and above the x-axis is 1/(3 ln3 2).
144 CHAPTER 5 Indeterminate Forms

EXAMPLE 5.25
Because of in¬‚ation, the value of a dollar decreases as time goes on. Indeed,
this decrease in the value of money is directly related to the continuous
compounding of interest. For if one dollar today is invested at 6% contin-
uously compounded interest for ten years then that dollar will have grown
to e0.06·10 = $1.82 (see Section 6.5 for more detail on this matter). This
means that a dollar in the currency of ten years from now corresponds to
only e’0.06·10 = $0.55 in today™s currency.
Now suppose that a trust is established in your name which pays 2t + 50
dollars per year for every year in perpetuity, where t is time measured in
years (here the present corresponds to time t = 0). Assume a constant
interest rate of 6%, and that all interest is re-invested. What is the total
value, in today™s dollars, of all the money that will ever be earned by your
trust account?

SOLUTION
Over a short time increment [tj ’1 , tj ], the value in today™s currency of the
money earned is about

(2tj + 50) · e’0.06·tj · tj .

The corresponding sum over time increments is

(2tj + 50) · e’0.06·tj tj .
j

This in turn is a Riemann sum for the integral

(2t + 50)e’0.06t dt.

If we want to calculate the value in today™s dollars of all the money earned from
now on, in perpetuity, this would be the value of the improper integral
+∞
(2t + 50)e’0.06t dt.
0

This value is easily calculated to be $1388.89, rounded to the nearest cent.

You Try It: A trust is established in your name which pays t + 10 dollars per year
for every year in perpetuity, where t is time measured in years (here the present
corresponds to time t = 0). Assume a constant interest rate of 4%. What is the total
value, in today™s dollars, of all the money that will ever be earned by your trust
account?
145
CHAPTER 5 Indeterminate Forms

Exercises
1. If possible, use l™Hôpital™s Rule to evaluate each of the following limits. In
each case, check carefully that the hypotheses of l™Hôpital™s Rule apply.
cos x ’ 1
(a) lim
x’0 x 2 ’ x 3
e2x ’ 1 ’ 2x
(b) lim
x2 + x4
x’0
cos x
(c) lim
x’0 x 2
[ln x]2
(d) lim
x’1 (x ’ 1)
(x ’ 2)3
(e) lim
x’2 sin(x ’ 2) ’ (x ’ 2)
ex ’ 1
(f ) lim
x’1 x ’ 1
2. If possible, use l™Hôpital™s Rule to evaluate each of the following limits. In
each case, check carefully that the hypotheses of l™Hôpital™s Rule apply.
x3
(a) lim
x’+∞ ex ’ x 2
ln x
(b) lim
x’+∞ x
e’x
(c) lim
x’+∞ ln[x/(x + 1)]

sin x
(d) lim
x’+∞ e’x
ex
(e) lim
x’’∞ 1/x

ln |x|
(f ) lim
x’’∞ e’x
3. If possible, use some algebraic manipulations, plus l™Hôpital™s Rule, to
evaluate each of the following limits. In each case, check carefully that the
hypotheses of l™Hôpital™s Rule apply.
lim x 3 e’x
(a)
x’+∞
lim x · sin[1/x]
(b)
x’+∞
146 CHAPTER 5 Indeterminate Forms

lim ln[x/(x + 1)] · (x + 1)
(c)
x’+∞
lim ln x · e’x
(d)
x’+∞
lim e2x · x 2
(e)
x’’∞
lim x · e1/x
(f )
x’0
4. Evaluate each of the following improper integrals. In each case, be sure to
write the integral as an appropriate limit.
1
x ’3/4 dx
(a)
0
3
(x ’ 3)’4/3 dx
(b)
1
2 1
(c) dx
(x + 1)1/3
’2
6 x
(d) dx
(x ’ 1)(x + 2)
’4
x+4
8
(e) dx
(x ’ 8)1/3
4
3 sin x
(f ) dx
x2
0
5. Evaluate each of the following improper integrals. In each case, be sure to
write the integral as an appropriate limit.

e’3x dx
(a)
1

x 2 e’x dx
(b)
2

(c) x ln x dx
0
∞ dx
(d)
1 + x2
1
∞ dx
(e)
x
1
’1 dx
(f )
x2 + x
’∞
CHAPTER 6



Transcendental
Functions
6.0 Introductory Remarks
There are two types of functions: polynomial and transcendental. A polynomial
of degree k is a function of the form p(x) = a0 + a1 x + a2 x 2 + · · · + ak x k .
Such a polynomial has precisely k roots, and there are algorithms that enable us to
solve for those roots. For most purposes, polynomials are the most accessible and
easy-to-understand functions. But there are other functions that are important in
mathematics and physics. These are the transcendental functions. Among this more
sophisticated type of functions are sine, cosine, the other trigonometric functions,
and also the logarithm and the exponential. The present chapter is devoted to the
study of transcendental functions.



6.1 Logarithm Basics
A convenient, intuitive way to think about the logarithm function is as the inverse
to the exponentiation function. Proceeding intuitively, let us consider the function
f (x) = 3x .
To operate with this f, we choose an x and take 3 to the power x. For example,
f (4) = 34 = 3 · 3 · 3 · 3 = 81

147
Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
148 CHAPTER 6 Transcendental Functions

1
f (’2) = 3’2 =
9
f (0) = 30 = 1.
The inverse of the function f is the function g which assigns to x the power to
which you need to raise 3 to obtain x. For instance,
g(9) = 2 because f (2) = 9
g(1/27) = ’3 because f (’3) = 1/27
g(1) = 0 because f (0) = 1.
We usually call the function g the “logarithm to the base 3” and we write g(x) =
log3 x. Logarithms to other bases are de¬ned similarly.
While this approach to logarithms has heuristic appeal, it has many drawbacks:
we do not really know what 3x means when x is not a rational number; we have
no way to determine the derivative of f or of g; we have no way to determine the
integral of f or of g. Because of these dif¬culties, we are going to use an entirely


Y
new method for studying logarithms. It turns out to be equivalent to the intuitive
FL
method described above, and leads rapidly to the calculus results that we need.

6.1.1 A NEW APPROACH TO LOGARITHMS
AM

When you studied logarithms in the past you learned the formula
log(x · y) = log x + log y;
TE



this says that logs convert multiplication to addition. It turns out that this property
alone uniquely determines the logarithm function.
Let (x) be a differentiable function with domain the positive real numbers
and whose derivative function (x) is continuous. Assume that satis¬es the
multiplicative law
(x · y) = (x) + (y) (—)
for all positive x and y. Then it must be that (1) = 0 and there is a constant C
such that
C
(x) = .
x
In other words
xC
(x) = dt.
t
1
A function (x) that satis¬es these properties is called a logarithm function.
The particular logarithm function which satis¬es (1) = 1 is called the natural
149
CHAPTER 6 Transcendental Functions

logarithm: In other words,
x 1
natural logarithm = ln x = dt.
t
1
For 0 < x < 1 the value of ln x is the negative of the actual area between the
graph and the x-axis. This is so because the limits of integration, x and 1, occur in
x
reverse order: ln x = 1 (1/t) dt with x < 1.




Fig. 6.1

Notice the following simple properties of ln x which can be determined from
looking at Fig. 6.1:
(i) When x > 1, ln x > 0 (after all, ln x is an area).
(ii) When x = 1, ln x = 0.
(iii) When 0 < x < 1, ln x < 0
1
x 1 1
dt = ’
since dt < 0 .
t t
1 x

(iv) If 0 < x1 < x2 then ln x1 < ln x2 .
We already know that the logarithm satis¬es the multiplicative property. By
applying this property repeatedly, we obtain that: If x > 0 and n is any integer then
ln(x n ) = n · ln x.
A companion result is the division rule: If a and b are positive numbers then
a
= ln a ’ ln b.
ln
b
EXAMPLE 6.1
Simplify the expression
a 3 · b2
A = ln .
c’4 · d
150 CHAPTER 6 Transcendental Functions

SOLUTION
We can write A in simpler terms by using the multiplicative and quotient
properties:
A = ln(a 3 · b2 ) ’ ln(c’4 · d)
= [ln a 3 + ln(b2 )] ’ [ln(c’4 ) + ln d]
= [3 ln a + 2 · ln b] ’ [(’4) · ln c + ln d]
= 3 ln a + 2 · ln b + 4 · ln c ’ ln d.
The last basic property of the logarithm is the reciprocal law: For any x > 0
we have
ln(1/x) = ’ ln x.
EXAMPLE 6.2
Express ln(1/7) in terms of ln 7. Express ln(9/5) in terms of ln 3 and ln 5.

SOLUTION
We calculate that
ln(1/7) = ’ ln 7,
ln(9/5) = ln 9 ’ ln 5 = ln 32 ’ ln 5 = 2 ln 3 ’ ln 5.

You Try It: Simplify ln(a 2 b’3 /c5 ).

6.1.2 THE LOGARITHM FUNCTION AND THE
DERIVATIVE
Now you will see why our new de¬nition of logarithm is so convenient. If we want
to differentiate the logarithm function we can apply the Fundamental Theorem of
Calculus:
x 1 1
d d
ln x = dt = .
dx dx t x
1
More generally,
1 du
d
ln u = .
dx u dx
EXAMPLE 6.3
Calculate
d d d d d
ln(4 + x), ln(x 3 ’ x), [(ln x)5 ], [(ln x) · (cot x)].
ln(cos x),
dx dx dx dx dx
151
CHAPTER 6 Transcendental Functions

SOLUTION
For the ¬rst problem, we let u = 4 + x and du/dx = 1. Therefore we have

1 1
d d
ln(4 + x) = · (4 + x) = .
4 + x dx 4+x
dx
Similarly,

3x 2 ’ 1
1
d d3
ln(x ’ x) = 3 · (x ’ x) = 3
3
x ’ x dx x ’x
dx
’ sin x
1
d d
ln(cos x) = · (cos x) =
cos x dx cos x
dx
5(ln x)4
41
d d
[(ln x) ] = 5(ln x) · (ln x) = 5(ln x) · =
5 4
dx dx x x
d d d
[(ln x) · (cot x)] = ln x · (cot x) + (ln x) · cot x
dx dx dx
1
= · cot x + (ln x) · (’ csc2 x).
x

You Try It: What is the derivative of the function ln(x 3 + x 2 ) ?

Now we examine the graph of y = ln x. Since
1
d
(ln x) =
(i) > 0,
dx x
d2 d1 1
(ln x) = =’
(ii) < 0,
x2
dx dx x
(iii) ln(1) = 0,
we know that ln x is an increasing, concave down function whose graph passes
through (1, 0). There are no relative maxima or minima (since the derivative is
never 0). Certainly ln .9 < 0; the formula ln(.9n ) = n ln .9 therefore tells us that
ln x is negative without bound as x ’ 0+ . Since ln x = ’ ln(1/x), we may also
conclude that ln x is positive without bound as x ’ +∞. A sketch of the graph of
y = ln x appears in Fig. 6.2.
We learned in the last paragraph that the function ln x takes negative values which
are arbitrarily large in absolute value when x is small and positive. In particular,
the negative y axis is a vertical asymptote. Since ln(1/x) = ’ ln x, we then ¬nd
that ln x takes arbitrarily large positive values when x is large and positive. The
graph exhibits these features.
Since we have only de¬ned the function ln x when x > 0, the graph is only
sketched in Fig. 6.2 to the right of the y-axis. However it certainly makes sense to
152 CHAPTER 6 Transcendental Functions




Fig. 6.2

discuss the function ln |x| when x = 0 (Fig. 6.3):




Fig. 6.3

If x = 0 then
1
d
(ln |x|) = .
dx x
In other words,
1
dx = ln |x| + C.
x
More generally, we have
1 du
d
ln |u| =
dx u dx
and
1 du
dx = ln |u| + C.
u dx
EXAMPLE 6.4
Calculate
4 1
dx, dx.
x+1 ’2 + 3x
153
CHAPTER 6 Transcendental Functions

SOLUTION

4 1
dx = 4 dx = 4 ln |x + 1| + C
x+1 x+1
1 1
dx = ln | ’ 2 + 3x| + C.
’2 + 3x 3
You Try It: Calculate the integral
cos x
dx.
2 + sin x
You Try It: Calculate the integral
e 1
dx.
x · [ln x]3/2
1
EXAMPLE 6.5
Evaluate the integral
cos x
dx.
3 sin x ’ 4
SOLUTION
For clarity we set •(x) = 3 sin x ’ 4, • (x) = 3(cos x). The integral then
has the form
1 1
• (x)
dx = ln |•(x)| + C.
3 3
•(x)
Resubstituting the expression for •(x) yields that
cos x 1
dx = ln |3 sin x ’ 4| + C.
3 sin x ’ 4 3
x2
You Try It: Evaluate dx.
1 ’ x3
EXAMPLE 6.6
Calculate

cot x dx.

SOLUTION
We rewrite the integral as
cos x
dx.
sin x
154 CHAPTER 6 Transcendental Functions

For clarity we take •(x) = sin x, • (x) = cos x. Then the integral becomes
• (x)
dx = ln |•(x)| + C.
•(x)
Resubstituting the expression for • yields the solution:

cot x dx = ln | sin x| + C.




6.2 Exponential Basics
Examine Fig. 6.4, which shows the graph of the function
f (x) = ln x, x > 0.




Fig. 6.4

As we observed in Section 1, the function f takes on all real values. We already
have noticed that, since
1
d
ln x = > 0,
dx x
the function ln x is increasing. As a result,
ln : {x : x > 0} ’ R
is one-to-one and onto. Hence the natural logarithm function has an inverse.
155
CHAPTER 6 Transcendental Functions

The inverse function to the natural logarithm function is called the exponential
function and is written exp(x). The domain of exp is the entire real line. The range
is the set of positive real numbers.

EXAMPLE 6.7
Using the de¬nition of the exponential function, simplify the expressions

exp(ln a + ln b) ln(7 · [exp(c)]).
and

SOLUTION
We use the key property that the exponential function is the inverse of the
logarithm function. We have
exp(ln a + ln b) = exp(ln(a · b)) = a · b,
ln(7 · [exp(c)]) = ln 7 + ln(exp(c)) = ln 7 + c.

You Try It: Simplify the expression ln(a 3 · 35 · 5’4 ).

6.2.1 FACTS ABOUT THE EXPONENTIAL FUNCTION
First review the properties of inverse functions that we learned in Subsection 1.8.5.
The graph of exp(x) is obtained by re¬‚ecting the graph of ln x in the line y = x.
We exhibit the graph of y = exp(x) in Fig. 6.5.




Fig. 6.5

We see, from inspection of this ¬gure, that exp(x) is increasing and is concave
up. Since ln(1) = 0 we may conclude that exp(0) = 1. Next we turn to some of
the algebraic properties of the exponential function.
156 CHAPTER 6 Transcendental Functions

For all real numbers a and b we have
(a) exp(a + b) = [exp(a)] · [exp(b)].
exp(a)
(b) For any a and b we have exp(a ’ b) = .
exp(b)
These properties are veri¬ed just by exploiting the fact that the exponential is
the inverse of the logarithm, as we saw in Example 6.7.
EXAMPLE 6.8
Use the basic properties to simplify the expression

[exp(a)]2 · [exp(b)]3
.
[exp(c)]4
SOLUTION
We calculate that

[exp(a)]2 · [exp(b)]3 [exp(a)] · [exp(a)] · [exp(b)] · [exp(b)] · [exp(b)]
=
[exp(c)] · [exp(c)] · [exp(c)] · [exp(c)]
[exp(c)]4
exp(a + a + b + b + b)
=
exp(c + c + c + c)
= exp(a + a + b + b + b ’ c ’ c ’ c ’ c)
= exp(2a + 3b ’ 4c).

You Try It: Simplify the expression (exp a)’3 · (exp b)2 / exp(5c).

6.2.2 CALCULUS PROPERTIES OF THE EXPONENTIAL
Now we want to learn some “calculus properties” of our new function exp(x).
These are derived from the standard formula for the derivative of an inverse, as in
Section 2.5.1.
For all x we have
d
(exp(x)) = exp(x).
dx
In other words,

exp(x) dx = exp(x).

More generally,
d du
exp(u) = exp(u)
dx dx
157
CHAPTER 6 Transcendental Functions

and
du
dx = exp(u) + C.
exp(u)
dx
We note for the record that the exponential function is the only function (up to
constant multiples) that is its own derivative. This fact will come up later in our
applications of the exponential
EXAMPLE 6.9
Compute the derivatives:
d d d
([exp(x)] · [cot x]).
exp(4x), (exp(cos x)),
dx dx dx
SOLUTION
For the ¬rst problem, notice that u = 4x hence du/dx = 4. Therefore we
have
d d
exp(4x) = [exp(4x)] · (4x) = 4 · exp(4x).
dx dx
Similarly,
d d
(exp(cosx)) = [exp(cosx)]· cosx = [exp(cosx)]·(’sin x),
dx dx
d d d
([exp(x)]·[cot x]) = exp(x) ·(cot x)+[exp(x)]· cot x
dx dx dx
= [exp(x)]·(cot x)+[exp(x)]·(’csc2 x).
You Try It: Calculate (d/dx)(exp(x · sin x)).
EXAMPLE 6.10
Calculate the integrals:

[exp(x)]3 dx, exp(2x + 7) dx.
exp(5x) dx,

SOLUTION
We have
1
exp(5x) dx = exp(5x) + C
5
[exp(x)]3 dx = [exp(x)] · [exp(x)] · [exp(x)] dx
1
= exp(3x) dx = exp(3x) + C
3
1 1
exp(2x + 7) dx = exp(2x + 7) · 2 dx = exp(2x + 7) + C.
2 2
158 CHAPTER 6 Transcendental Functions

EXAMPLE 6.11
Evaluate the integral

[exp(cos3 x)] · sin x · cos2 x dx.

SOLUTION
For clarity, we let •(x) = cos3 x, • (x) = 3 cos2 x · (’ sin x). Then the
integral becomes
1 1
’ exp(•(x)) · • (x) dx = ’ exp(•(x)) + C.
3 3
Resubstituting the expression for •(x) then gives
1
[exp(cos3 x)] · sin x · cos2 x dx = ’ exp(cos3 x) + C.
3
EXAMPLE 6.12



Y
Evaluate the integral
exp(x) + exp(’x)
FL
dx.
exp(x) ’ exp(’x)
AM

SOLUTION
For clarity, we set •(x) = exp(x) ’ exp(’x), • (x) = exp(x) + exp(’x).
Then our integral becomes
TE



• (x) dx
= ln |•(x)| + C.
•(x)
Resubstituting the expression for •(x) gives
exp(x) + exp(’x)
dx = ln | exp(x) ’ exp(’x)| + C.
exp(x) ’ exp(’x)
x · exp(x 2 ’ 3) dx.
You Try It: Calculate

6.2.3 THE NUMBER e
The number exp(1) is a special constant which arises in many mathematical and
physical contexts. It is denoted by the symbol e in honor of the Swiss mathematician
Leonhard Euler (1707“1783) who ¬rst studied this constant. We next see how to
calculate the decimal expansion for the number e.
In fact, as can be proved in a more advanced course, Euler™s constant e satis¬es
the identity
n
1
1+ = e.
lim
n
n’+∞
159
CHAPTER 6 Transcendental Functions

[Refer to the “You Try It” following Example 5.9 in Subsection 5.2.3 for a
consideration of this limit.]
This formula tells us that, for large values of n, the expression
n
1
1+
n
gives a good approximation to the value of e. Use your calculator or computer to
check that the following calculations are correct:
n
1
n = 10 1+ = 2.5937424601
n
n
1
n = 50 1+ = 2.69158802907
n
n
1
n = 100 1+ = 2.70481382942
n
n
1
n = 1000 1+ = 2.71692393224
n
n
1
n = 10000000 1+ = 2.71828169254.
n
With the use of a suf¬ciently large value of n, together with estimates for the error
term
n
1
e’ 1+ ,
n
it can be determined that
e = 2.71828182846
to eleven place decimal accuracy. Like the number π, the number e is an irrational
number. Notice that, since exp(1) = e, we also know that ln e = 1.
EXAMPLE 6.13
Simplify the expression
ln(e5 · 8’3 ).

SOLUTION
We calculate that
ln(e5 · 8’3 ) = ln(e5 ) + ln(8’3 )
= 5 ln(e) ’ 3 ln 8
= 5 ’ 3 ln 8.
160 CHAPTER 6 Transcendental Functions

You Try It: Use your calculator to compute log10 e and ln 10 = loge 10 (see
Example 6.20 below). Con¬rm that these numbers are reciprocals of each other.




6.3 Exponentials with Arbitrary Bases
6.3.1 ARBITRARY POWERS
We know how to de¬ne integer powers of real numbers. For instance
1 1
9’3 =
64 = 6 · 6 · 6 · 6 = 1296 =
and .
9·9·9 729
But what does it mean to calculate
4π or π e ?
You can calculate values for these numbers by punching suitable buttons on your
calculator, but that does not explain what the numbers mean or how the calcu-
lator was programmed to calculate them. We will use our understanding of the
exponential and logarithm functions to now de¬ne these exponential expressions.
If a > 0 and b is any real number then we de¬ne
a b = exp(b · ln a). (—)
To come to grips with this rather abstract formulation, we begin to examine some
properties of this new notion of exponentiation:
If a is a positive number and b is any real number then
(1) ln(a b ) = b · ln a.
In fact
ln(a b ) = ln(exp(b · ln a)).
But ln and exp are inverse, so that the last expression simpli¬es to b · ln a.

EXAMPLE 6.14
Let a > 0. Compare the new de¬nition of a 4 with the more elementary
de¬nition of a 4 in terms of multiplying a by itself four times.

SOLUTION
We ordinarily think of a 4 as meaning
a · a · a · a.
161
CHAPTER 6 Transcendental Functions

According to our new de¬nition of a b we have
a 4 = exp(4 · ln a) = exp(ln a + ln a + ln a + ln a)
= exp(ln[a · a · a · a]) = a · a · a · a.
It is reassuring to see that our new de¬nition of exponentiation is consistent
with the familiar notion for integer exponents.
EXAMPLE 6.15
Express exp(x) as a power of e.
SOLUTION
According to our de¬nition,
ex = exp(x · ln(e)).
But we learned in the last section that ln(e) = 1. As a result,
ex = exp(x).
You Try It: Simplify the expression ln[ex · x e ].
Because of this last example we will not in the future write the exponential
function as exp(x) but will use the more common notation ex . Thus
exp(ln x) = x becomes eln x = x
ln(exp(x)) = x becomes ln(ex ) = x
exp(a + b) = [exp(a)] · [exp(b)] becomes ea+b = ea eb
ea
exp(a)
exp(a ’ b) = =b
a’b
becomes e
exp(b) e
a = exp(b · ln a) becomes a = e
b b b·ln a
.
EXAMPLE 6.16
Use our new de¬nitions to simplify the expression
A = e[5·ln 2’3·ln 4] .
SOLUTION
We write
eln(32) 32 1
[ln(25 )’ln(43 )]
A=e =e = ln(64) = =.
ln 32’ln 64
64 2
e

We next see that our new notion of exponentiation satis¬es certain familiar rules.
If a, d > 0 and b, c ∈ R then
(i) a b+c = a b · a c
162 CHAPTER 6 Transcendental Functions

ab
=c
b’c
(ii) a
a
b )c = a b·c
(iii) (a
a b = d if and only if d 1/b = a (provided b = 0)
(iv)
a0 = 1
(v)
a1 = a
(vi)
(a · d)c = a c · d c .
(vii)

EXAMPLE 6.17
Simplify each of the expressions

5’7 · π 4
(32 · x 3 )4 .
4 ln 3
(e ) , ,
5’3 · π 2


SOLUTION
We calculate:

(e4 )ln 3 = e4·ln 3 = (eln 3 )4 = 34 = 81;
5’7 · π 4 1
= 5’7’(’3) · π 4’2 = 5’4 · π 2 = · π 2;
5’3 · π 2 625
(3 · x ) = (3 ) · (x ) = 3 · x = 6561 · x 12 .
2 34 24 34 8 12


You Try It: Simplify the expression ln[e3x · e’y’5 · 24 ].

EXAMPLE 6.18
Solve the equation
(x 3 · 5)8 = 9

for x.

SOLUTION
We have

(x 3 · 5)8 = 9 ’ x 3 · 5 = 91/8 ’ x 3 = 91/8 · 5’1
91/24
’1 1/3
’ x = (9 ·5 ’ x = 1/3 .
1/8
)
5

You Try It: Solve the equation 4x · 32x = 7. [Hint: Take the logarithm of both
sides. See also Example 6.22 below.]
163
CHAPTER 6 Transcendental Functions

6.3.2 LOGARITHMS WITH ARBITRARY BASES
If you review the ¬rst few paragraphs of Section 1, you will ¬nd an intuitively
appealing de¬nition of the logarithm to the base 2:
log2 x is the power to which you need to raise 2 to obtain x.
With this intuitive notion we readily see that
log2 16 = “the power to which we raise 2 to obtain 16” = 4
and
log2 (1/4) = “the power to which we raise 2 to obtain 1/4” = ’2.
However this intuitive approach does not work so well if we want to take logπ 5

or log2 7. Therefore we will give a new de¬nition of the logarithm to any base
a > 0 which in simple cases coincides with the intuitive notion of logarithm.
If a > 0 and b > 0 then
ln b
loga b = .
ln a
EXAMPLE 6.19
Calculate log2 32.

SOLUTION
We see that
5 · ln 2
ln 25
ln 32
log2 32 = = = = 5.
ln 2 ln 2 ln 2
Notice that, in this example, the new de¬nition of log2 32 agrees with the
intuitive notion just discussed.
EXAMPLE 6.20
Express ln x as the logarithm to some base.

SOLUTION
If x > 0 then
ln x ln x
loge x = = = ln x.
ln e 1
Thus we see that the natural logarithm ln x is precisely the same as loge x.

Math Note: In mathematics, it is common to write ln x rather than loge x.

You Try It: Calculate log3 27 + log5 (1/25) ’ log2 8.
164 CHAPTER 6 Transcendental Functions

We will be able to do calculations much more easily if we learn some simple
properties of logarithms and exponentials.
If a > 0 and b > 0 then
a (loga b) = b.
If a > 0 and b ∈ R is arbitrary then
loga (a b ) = b.
If a > 0, b > 0, and c > 0 then
(i) loga (b · c) = loga b + loga c
(ii) loga (b/c) = loga b ’ loga c
logc b
(iii) loga b =
logc a
1
(iv) loga b =
logb a
(v) loga 1 = 0
(vi) loga a = 1
(vii) For any exponent ±, loga (b± ) = ± · (loga b)
We next give several examples to familiarize you with logarithmic and
exponential operations.

EXAMPLE 6.21
Simplify the expression

log3 81 ’ 5 · log2 8 ’ 3 · ln(e4 ).

SOLUTION
The expression equals

log3 (34 )’5·log2 (23 )’3·ln e4 = 4·log3 3’5·[3·log2 2]’3·[4·ln e]
= 4·1’5·3·1’3·4·1 = ’23.

You Try It: What does log3 5 mean in terms of natural logarithms?

EXAMPLE 6.22
Solve the equation
4
5x · 23x =
7x
for the unknown x.
165
CHAPTER 6 Transcendental Functions

SOLUTION
We take the natural logarithm of both sides:
4
ln(5x · 23x ) = ln .
7x
Applying the rules for logarithms we obtain
ln(5x ) + ln(23x ) = ln 4 ’ ln(7x )
or
x · ln 5 + 3x · ln 2 = ln 4 ’ x · ln 7.
Gathering together all the terms involving x yields
x · [ln 5 + 3 · ln 2 + ln 7] = ln 4
or
x · [ln(5 · 23 · 7)] = ln 4.
Solving for x gives
ln 4
x= = log280 4.
ln 280
EXAMPLE 6.23
Simplify the expression
5 · log7 3 ’ (1/4) · log7 16
B= .
3 · log7 5 + (1/5) · log7 32

SOLUTION
The numerator of B equals

log7 (35 ) ’ log7 (161/4 ) = log7 243 ’ log7 2 = log7 (243/2).
Similarly, the denominator can be rewritten as
log7 53 + log7 (321/5 ) = log7 125 + log7 2 = log7 (125 · 2) = log7 250.
Putting these two results together, we ¬nd that
log7 243/2
B= = log250 (243/2).
log7 250

32
You Try It: What does mean (in terms of the natural logarithm function)?
EXAMPLE 6.24
Simplify the expression (log4 9) · (log9 16).

<<

. 5
( 11)



>>