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.