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12. EXPONENTIAL AND LOGARITHMIC FUNCTION Exponential and logarithmic functions have many important applications in engineering, biology, chemistry, geology, and many other fields. In this section we will define exponential and logarithmic functions and study their properties. In particular, we shall see that these functions are inverses of each other. Exponential function Suppose that b is any fixed positive real number, and suppose R is the set of real numbers. DEFNITION if b > 0 and x e R is any real number, then the function defined by is called the EXPONENTIAL FUNCTION with base b . The DOMABI of f (x) = b x is the set of all real numbers. The RANGE of the exponential function is the set of all POSITIVE real numbers . We know the laws of exponents for rational exponents. ( see Chapter 5 : Exponents and their properties.) We recall that if b > 0 and c and d are arbifrary rational numbers, then b c .b d —b C + d = bC-d b 0 =1 1 b —c With the definition above the Laws ofExponents hold for all REAL EXPONENTS. ( It will be proved next year . ) 110

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12. EXPONENTIAL AND LOGARITHMIC FUNCTION

Exponential and logarithmic functions have many important applications in engineering,

biology, chemistry, geology, and many other fields. In this section we will define

exponential and logarithmic functions and study their properties. In particular, we shall

see that these functions are inverses of each other.

Exponential function

Suppose that b is any fixed positive real number, and suppose R is the set of real

numbers.

DEFNITION

if b > 0 and x e R is any real number, then the function defined by

is called the EXPONENTIAL FUNCTION with base b .

The DOMABI of f (x) = b x is the set of all real numbers.

The RANGE of the exponential function is the set of all POSITIVE real numbers .

We know the laws of exponents for rational exponents. ( see Chapter 5 : Exponents and

their properties.)

We recall that if b > 0 and c and d are arbifrary rational numbers, then

b c .b d —b C + d — = bC-d

b 0 =1

1

b—c

With the definition above the Laws ofExponents hold for all REAL EXPONENTS.

( It will be proved next year .)

110

EXAMPLE 1

Sketch the graph of the exponential function

SOLUTION : Let us make a brief table of values :

x101112131 I-h 1-1/3 11Y 4181 h 11/81

9

8

7

6

5

4

3

2

1

Now we list several important facts regarding this function and its graph.

l.) As x becomes very large and positive, f (x) = 2 x also becomes very large andpositive. We say that f (x) grows without bound as x grows without bound.

2.) As x grows numerically large in the negative direction, f (x) = 2 x approaches 0.

3.) Since 2 | , the graph crosses the y axis at the point (0 , 1).

4.) f (x) = 2 x is strictly increasing on its entire domain, making it a one-to-one

function.

This graph is typical ofexponential functions WITH BASE GREATER THAN 1 .

1.) Since 1 , each ofthese graphs has (0 , 1) as its Y-INTERCEPT .

2.) For each of these graphs, as x —+ + , b x rapidly becomes extremely large :

f(x) = b x — +Ø . This behavior is known as EXPONENTIAL GROWTH .

3.) Finally, for each of these graphs, as x , f (x) = b x--»0 (with b > I

That is, the x axis is a HORIZONTAL ASYVTTOTE ofthese graphs.

111

EXAMPLE 2

Sketch the graph of the exponential function

(1/2)x.

SOLUTION : Comparing the function

with the function f (x) 2 x , we note that the image of x under f (x) = (1/2) x will beexactly the same as the image of ( - x ) under the function 2 x .

Hence, the graph ofthese THE Y AXIS.

are REFLECTIONS OF EACH OTHER ACROSS

9

8

6

5

4

3

1

1 2 3 4

Now we list the properties of the graph :

1.) As x becomes very large and positive, f (x) = (1/2) x becomes very close tozero.

2.) As x becomes very large in the negative direction, f (x) becomes very largeand positive.

3.) The y intercept point is (0 , l) .

4.) f (x) = (1/2) x is strictly decreasing on its entire domain, making it a one-to-onefunction.

This graph is typical ofexponential functions WITH BASE LESS THAN 1

112

NOTICE the PROPERTIES ofthe EXPONENTLAL FUNCTIONS

a.) If b > 1 , f (x) = b x is a strictly increasing function.

b.) If 0 < b < 1 , f (x) = b x is a strictly decreasing function.

c.) All exponential functions f (x) = b x , regardless of their base, have a

common point of intersection : P ( 0, 1 ) .Because

if x = 0 , then b 1 , for all values of b greater than zero.

EXERCISES

1. Sketch the graph of the functions :

a.) y=3X

c.) f (x) = (1/4) x

2. Tell which of the following statements are frue and which are false. Justify your

a.) b X - Y =b X -b Y b.) f (x) = b x is strictly decreasing

c.) f (x) = x 2 / 3 is an exponential function

d.) If c is a positive real constant and - oo < x < + , then f (x) = c x is strictly

increasing.

e.) The graph of f (x) = b x is symmetric about the y axis.

f.) The graph of f (x) = b x has no x— intercept point.

3. Determine which function is odd, even, or neither even nor odd :

d.) f (x) = x 2 +2 X

e.) f (x) =x.3 x

113

Logarithmic function

Since the exponential function is a one-to-one function, this function has an inverse.

DEFNITION :

If b > 0 and b 1 , then the function defined by

is called the LOGARITHMIC FUNCTION to the base b.

The logarithmic function to the base b is the inverse function of the exponential function

with base b .

That means,

y = log b x if and only if b Y =x.

This equality says that logarithms are exponents. Indeed

log b x is simply that exponent to which we should raise the base b in

order to obtain the positive number x .

For instance, log 2 8 is the exponent that the base 2 must be raised in order to obtain 8 ;that is ,

log 2 8 = 3 , since 2 3

Two important identities :

1.) y = log b x

2.) y=bX

if and only if

if and only if

b (log x)

log b (b X ) = x

EXERCISES

1. Put each equation in logarithmic form :

a.) 2 5 ̄ 32 .

c) 4 - = 1/64 d.)

2. Find each of the given logarithms :

a.) log 10 10 b.)

c.) log 0.0001 d.)

3. Find each of the given logarithms :

a.) log 82

c.) log 4 ( 1/2)

b.)

d.)

7 0 =1

= 81

log 100000

log 64 4

log 4 64

log 4 2

114

e.) log 16 16

g.) log 3 ( 1/81)

i.) log 8 32

k.) log 2 ( 1/16)

f.) log 5 25

h.) log 1/3 27

j.) log ( 1/27)

l.) log 4 1

4. Find the unknown N , b , u in each equation :

a.) log b 32=5

c.) log9N= l/2

e.) log 8 = u

g.) log b 1=0

i.) log 4N=-3

k.) log 5N=-2

m.) log b ( 1/7)=- 1/3

o.) log 4 (log 4 16) = u

r.) log 2 ( log

t.) 2 (log 3)

b.) log b 81 = 4

d.) log 8 N = 2/3

f.) log 32 8 = u

h.) log 55

j.) log b ( 1/9)

l.) log 3 81 = u

n.)log 681 = u

p.) log b ( log b 27 3 )

s.) log 4 N = 4

Since the logarithmic function is the inverse function of the exponential function ,

the DOMABI of the logarithmic function is ( 0 , + oo ) and

the RANGE of the logarithmic ftnction is ( - 00, + 00 )

Since they are inverse functions, the graph of the logarithmic function to the base b is thereflection across the line y = x of the exponential function with base b.

If b > 1 , y = b x is a strictly increasing function, hence

y = log b x is also a strictly increasing, function.

If O < b < 1 , y = b x is a strictly decreasing function, hence

y = log b x is also a strictly decreasing function.

115

EXAMPLE 3

Sketch the graph of

y = log 2 x and y = 2 x

(0,1)

on the same coordinate axes.

y=28

y=10g2x

(1,0)

Notice the properties of the logarithmic function :

a.) The x intercept point is ( 1 , 0) .

b.) The graph has no y intercept point, but the y axis is an asymptote of

the graph of y = log b x .

c.) All logarithmic functions f (x) = log b x , regardless of their base, havea common point of intersection. If x = 1 , then log b x = 0 , for all values of

b greater than zero.

116

EXAMPLE 4

Sketch on the same coordinate axes

y = log % x and y = (1/2) x

(0,1

Flog 1/2 X

EXERCISES

Sketch the graph of the functions

a.) y = 3 x and y = log 3 x

and y = log 1/5 (x)

117

Laws of logarithms

By using the definition of logarithms and the laws of exponents, three properties of

logarithms will be proved •

LAW 1.

The logarithm of the product of two positive numbers is equal to the sum of

the logarithms of the two numbers.

log b a.c = log b a + log b c

LAW 2.

The logarithm of the quotient of two positive numbers is equal to the

difference of logarithms of the two numbers.

alog b — = log b a— log b c

LAW 3.

The logarithm of a power of a positive number is equal to the exponent

times the logarithm of the number.

log b ac = c.log b a

PROOF: To prove LAW 1., let

x = log b a and y = log b c.

But by definition these two equations can be written in exponential form :

a = b X and c = b Y , respectively. Then

While changing this to logarithmic form yields

log b a.c = X + y.

Since x = log b a and y = log b c , the equation above becomes

log b a.c = log b a + log b c,

which is the desired result.

Laws 2. and 3. are proved similarly.

EXAbåPLE 5

Find log 3 Tii

SOLUTION : By Law 3. we have log3 (27) 1 / 5 1/5) . log 327 .

Since log 3 27 = 3 , we have

1/5 .10g3 27 = 1/5 . 3 = 3/5 .

118

EXAMPLE 6

Assuming x , y and z are positive, express the following in terms of

logarithms of x , y and z .1/7

log b

1/7

SOLUTION : we have log b - (1/7) . log

— (1/7)[ log b y— ( 6.10g b x +4.10g b z)] = (1/7)log b y— (6/7)log b x— (4/7)log b z .

EXAMPLE 7

Solve the following equations :

a.) log 5 3.10g 5 15 —2.10g5 6+ log 5 4— log 5 25

b.) log 6 (3x— 1) — log 6 x = log 64

SOLUTION :

a.) The equation is equivalent to

log 5 (4x + 7) = log 5 15 3 —log 5 6 2 + log 5 4— log 5 25 . (see Law 3.)

Rewriting the equation (using Law 1. and 2.), we obtain

153.4log 5 (4x + 7) = log 5

36.25

hence 4x 15

Checking the solution in the original equation shows that x = 2 is a solution.

b.) log 6 (3x — 1) — log 6 x = log 64

Rewriting the equation using Law 1. and 2. ,we obtain

3x-1log 6 = log 64,

x

3x-1hence

x

We obtain 1

Hence,

This value does not satisfy the original equation since log ( -1) does not exist. So the

equation has no solution.

119

EXERCISES

Solve the following equations :

1. log 7 (3x —5) = 2.10g7 12 -- log 79

2. log 2 (x + 8) — log 2 (2x + 1) = log 8 2 + log 8 4

3. log 13 (x —3) + log 13 (x + 1) = log 13 5

4. 2.10g 2 5 + log 2 4= 2.10g 2 2x

5. log 11 (x + 1) = log 11 4— log Il (x— 2)

6. log 9 (x + 3) + log 9 (x —2) = 1/3.10g 9 8

7. log 10 x + log 10 (4x + 3) = 2.10g 10 5.Jä - log 10 5

8. log (x + I) + log 3 (x— l) = 1

9. h . log b X— log b 4 = log b 3

10. 2 -10.2 12

11. log 8 (x + 2) = log 8 (log 2 8) + log 8 (x—4)

12. 1/3 . log 20 (2x — 1) = log 20 11—2. log 20

13. log 6 (log 6 x) = 0

14. log 4 [ log 4 (log 4 x)] = 0

15. log 3 [ log 4 (log 5 x) ] = 0

16. Solve each system of equations :

a.) h . log 2 y— log 24

log 3 x + 102 3 2y=3

c.) log 2 [log 3 (x + y)]

log x + log y = 3.10g 2

e.) log (x +5)

3x +2y=30

120

+ log 5 ( log 7 7)

b.)

d.)

f.)

log 2 x + log 2 y = 5

log 2 x— log 2 y = 3

log x— h. log y = 0

x 2 —5 y 2 +4=0

log 7 (5x — 3y) = 0

9 x .3 X -

Y =27

Common and natural logarithms

Two bases are commonly employed for logarithms, 10 and e .

a.) Logarithms to the base 10 are called COMMON LOGARITHMS.

Common logarithms are denoted simply as log x or lg x.

That is

log x = lgx = log 10 x.

We can find the common logarithm of a number in a table of logarithm or on a

calculator.

For example,

log 87,6 = 1,9425...

The common logarithm of a number has two parts:

The integer part, called the CHARACTERISTIC of the logarithm. In our examplethe characteristic is 1 .

The other part is called the MANTISSA of the number. The mantissa is always lessthan 1 . In our example the mantissa is 0,9425...

If we are given log x , we can determine the ANTILOGARITI-n.'1 : x .

For example, Find x if log x = 0,444 ,

antilog = 2,78

b.) Logarithms to the base e are called NATURAL LOGARITffv1S .

They are given the special notation In x . That is

In x = log ex.

The base e is irrational , which correct to six places, is

e = 2,71828

Natural logarithms play an important role in both pure and applied mathematics.

Changing the base of a logarithm

Any positive number other than I may be used as a base of a system of logarithms.

A relationship between the logarithms of a number in two different bases will be

developed.

121

By the definition of logarithms

b log x

Taking the logarithm of both sides to base c , we have

log b x. log b = log c x

Thus, we have

log b x

For example,

log x =

logc x

logc b

orIn 10

log xInx =log e

These are the formulas to express natural logarithms in terms of common logarithms or toexpress common logarithms in terms of natural logarithms, respectively.

E)GRCISES

1. Solve the following equations :

a.) 3.10g 5 x + log 25 x = 7

b.) (log 3 x x ).( log 27 x) =4/3

c.) log 5 x 2 -2.10g 25 x = 3

d.) log 25 [ 1/5.10g3 (2— log 0.5 x =- 1/2

e.) log 3 log 2 x)

f.) log 4 x— log 0.25 x = 4

2. Find the domain of the functions :

log (x— 11) y = log (4 —x 2)

log (x — 2).(x — 3) y = log (x 4+16)

e.) Y = log1

y = logx +16

x 4g.) y=log = log (x2 —x 6)y x

log (x + 8) — log (x + Il) + log x

122

3. Solve the following equations :

a.) logs [4—2.10g 6 (5 —x) ] —1/3

b.) log 2

c.) log (4.5 125. (25 x I 1) -2

log(35 — x3 )d.)

log(5 — x)

e.) log (x— 1) + 1/3 = log 8 (x + 7)

f.) log 2+ log (4 +9) = I + log (2 x 2 +1)

g.) log (x + 3) (3X 2 + 2X + 9) = 2

h.) log 3 x— log x 1

4. Write in terms of simpler logarithmic form

a.c b a.blog + log — -2.10gc + log a— log

b

5. log (x) — log (y) = log (x + y) is ft-ue only if :

b.)

c.) x = 0 andy=l

2

6. Solve the equation :

2 6-3

2 —log3 x

123