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).