Pg219 [R] G1 5-36058 / HCG / Cannon & Elich cr 11-27-95 MP1 4.3 ...

4.3 Properties of Logarithmic Functions

219

with several values of c and draw graphs. Describe how the graphs change as c changes. (b) For what integer values of c do the graphs pass between P and Q?

56. P5, 4, Q5, 7

57. P5, 4, Q5, 7

Exercises 58?59 Your Choice From the family of functions f x cekx, where c and k are nonzero constants, choose c and k so that f satisfies the specified conditions.

58. The graphs of f and f 1 intersect in (a) QI (b) QIII. (c) Is there an f such that the graphs intersect in QIV? Explain.

59. (a) The graphs of f and y x 2 intersect in QIII and QIV.

(b) The graphs of f and y x intersect in QIII.

Exercises 60?61 (a) Find a formula for f. (b) If the graphs of f and f 1 intersect, find the point of intersection.

Use parametric mode.

60. f 1x log34 x

61. f 1x 24x 8

Exercises 62?64 Domains, Ranges, Graphs Function f

has an inverse. (a) Find the domain and range of f. (b) Find a formula for f 1 and give its domain and range.

62.

f x

1

1

3x

63.

f x

1

3x 3x

64.

f x

1

3x 3x

65. If f x 20.5x and gx Intx solve f gx 8. (Hint: Use dot mode in a decimal window.)

66. Repeat Exercise 65 for f gx 4.

Exercises 67?70 Graph to Formula Use the graph of y log2x to help match the function with one of the graphs a, b, c, or d. Think in terms of translations and reflections.

67. f x 1 log2x 69. f x log2x 1

68. f x log2x 70. f x log2x

y

y

4

3

2

1

? 2 ??11 ?2

(1, 0) x

1234

(2, ? 1)

?3

?4

(a)

4

3

2

(2, 2)

1 ? 2 ??11

(1, 1) x

1234

?2

?3

?4

(b)

y

4

3

2

(?1, 0) 1 ? 4 ? 3 ? 2 ??11

x 12

(? 2, ? 1) ? 2 ?3

?4

(c)

y

4 3 2 1

? 2 ??11 ?2 ?3 ?4

(2, 0) (3, 1) x

1234

(d)

4.3 P R O P E R T I E S O F L O G A R I T H M I C F U N C T I O N S

If one remembers . . . the useful concepts . . . as well as the countless misconceptions and errors that rigorous mathematical development avoids without touching, then mathematics begins to resemble not as much a nerve as the thread that Ariadne used to guide her lover Theseus out of the Labryinth in which he slew the dreaded Minotaur.

Hans C. von Baeyer

As defined in the preceding section, logarithms are exponents, so we would expect logarithms to have properties analogous to those of exponents. The list of some of the most important properties of both logarithms and exponents emphasizes the parallels between them.

220

Chapter 4 Exponential and Logarithmic Functions

Properties of logarithms and exponents

Logarithms

Exponents

L1. logbuv logb u logb v

u L2. logb v

logb u logb v

L3. logbu p plogb u

L4. logb 1 0 and logb b 1

E1. b ub v b uv

E2.

bu bv

b uv

E3. b up b up

E4. b0 1 and b1 b

Because logarithmic functions are defined only for positive numbers, L1, L2, and L3 are valid only when both u and v are positive.

Use the following equivalence statement to change a logarithmic equation to an exponential equation, and vice versa:

y logb x is equivalent to b y x.

(1)

I think it was during that

semester in Berkeley, when I was not quite fifteen, that I really switched into being serious about mathematics. As soon as I saw what geometry was about, it was immediately clear to me how the whole thing worked--I mean absolutely clear. I could visualize the figures rather well, and I didn't have any problem with understanding what proofs were supposed to be.

Andrew M. Gleason

We outline a proof of logarithm property L1; proofs for properties L2 and L3 are similar and are left as exercises (see Exercises 43 and 44). In words, property L1 states that the logarithm of a product is the sum of the logarithms.

logb u s and logb v t.

In terms of exponents

u b s and v b t.

Since the equation in property L1 involves uv, multiply the two exponential equations and apply exponent property E1 to get

uv b sb t b st.

Returning to logarithmic form,

logbuv s t.

Replacing s and t by logb u and logb v,

logbuv logb u logb v

The first three logarithm properties involve logarithms of products, quotients, and powers. We do not give similar formulas for sums and differences because there are no simple ways to express logbu v and logbu v in terms of logb u and logb v. Similarly, for exponents, we have for instance, b 2 ? b 3 b 5, but there is no simpler expression for b 2 b 3.

Strategy: Rewrite each ar-

gument as a power of the base: 81 34, 0.001 103, 1 312,

3

then use logarithm proper-

ties as appropriate.

EXAMPLE 1 Using logarithm properties Use properties L1 through L4 to evaluate

(a) log3 81 (b) log10 0.001 (c) log313.

Solution We indicate under the equals sign the property that gives the equality. Follow the strategy.

(a)

log3

81

log3

34

L3

4log3

3

L4

41

4,

hence

log3

81

4.

4.3 Properties of Logarithmic Functions

221

(b)

log10 0.001

log1010 3

L3

3log1010

L4

31

3,

hence log10 0.001 3.

(c)

log3

1 3

L2

log3

1

log3

3

L4

0

log3312

L3

1 2

log33

L4

1 2

,

hence

log3

1 3

1 2

EXAMPLE 2 Using logarithm properties Simplify.

(a) log5 10 log5 2 (b) log7 3 4log7 2

Solution

(a)

log5 10

log5 2

L2

log5

10 2

log5 5 1

L4

(b) log7 3 4log7 2 log7 3 log7 24 log7 3 log7 16

L3

log73 ? 16 log7 48

L1

It is important to learn to use logarithm properties L1 through L3 going from right to left, as well as left to right. For instance, in Example 2b we used property L3 to write

4log7 2 log7 24,

L3

and then we used property L1 to combine logarithms and get

log7 3 log7 16 log73 ? 16 log7 48.

L1

We call log7 48 a simplified form of log7 3 4log7 2. In a similar manner,

using L1 and L3 gives

log7 48 log73 ? 24 log7 3 log7 24 log7 3 4log7 2.

L1

L3

Thus, log7 48 can be written as a sum of logarithms, log7 3 4log7 2.

EXAMPLE 3 Combining logarithms Simplify.

logb x 4 logbx 1 logb 5.

Solution

logb x 4 logbx 1 logb 5 logb x logbx 14 logb 5

L3

logbxx 14 logb 5

L1

L2

logb

xx

5

14

Hence,

logb

x

4

logbx

1

logb

5

logb

xx

5

14

EXAMPLE 4 Numerical approximations In the next section we will show that four-place decimal approximations to log5 3 and log5 6 are:

log5 3 0.6826 log5 6 1.1133.

Use these values along with logarithm properties L1 through L4 to get threedecimal place approximations for (a) log5 2 and (b) log5 18.

222

Chapter 4 Exponential and Logarithmic Functions

Strategy Write each of 2

and 18 as a product, quo-

tient, power, etc. in terms of

the numbers 3 and 6, whose

logarithms

we

have:

2

6 3

,

18 3 ? 6, then use the

properties of logarithms.

Solution Follow the strategy.

(a)

log5

2

log5

6 3

log5

6

log5

3

1.1133 0.6826

0.4307.

Therefore, log5 2 0.431 to three decimal places. (b) log5 18 log53 ? 6 log5 3 log5 6

0.6826 1.1133

1.7959.

Hence, log5 18 1.796.

Strategy Combine the terms on the left by using property L2 to get a single logarithm equal to a constant, then express the result in exponential form. Since property L2 applies only to positive numbers, check all results in the original equation.

EXAMPLE 5 Using properties to solve equations Solve

(a)

log4 x

log4x

1

1 2

(b)

log4 x

log4x

1

1 2

.

Solution

(a) Follow the strategy,

log4

x

log4x

1

L2

log4

x x1

so the given equation can be written as

x

log4 x 1

1, 2

x

x

1

412,

or

x

x

1

2.

Solving for x, we find that x 2. Since log42 and log42 1 are both defined, we know that 2 belongs to the replacement set for the original equation and therefore 2 is the desired solution.

(b) As in part (a), the given equation can be written as

x

log4 x 1

1 ,

2

x

x

1

412,

or

x

x

1

2.

In this case, when we solve for x we find x 2. However, if we replace x with 2, the left side involves log42 and log42 1, neither of which is defined. Since 2 is not in the replacement set for the original equation, it

cannot be the solution. The given equation has no solution.

Example 5 illustrates an important point. Properties L1, L2, and L3 are valid for only positive values of all arguments; logarithmic functions are defined for only positive arguments. We could check the domains at each step, but it is good enough to check the final result in the original equation.

EXAMPLE 6 A logarithmic equation Solve the equation 2 log9 x 2 log9x 2 1.

4.3 Properties of Logarithmic Functions

223

Solution Divide through by 2 and write the left side in simpler form:

log9

x

log9 x

2

1 ,

2

log9xx

2

1 .

2

In exponential form,

xx 2 912, x 2 2x 3, or x 2 2x 3 0.

Solutions to the quadratic equation are 1 and 3. Since the domain of the original equation is the set of positive numbers, 1 is a solution but 3 is not.

Strategy: There is no for-

mula to simplify the loga-

rithm of a sum, rewrite 8x 8x as a power of 2 (the base) and then simplify.

EXAMPLE 7 Logarithms and sums Find the solution set for

(a) log28x 8x x 1 (c) log28x 8x 3x.

(b) log28x 8x 3x 1

Solution To simplify all three equations, begin with the expression 8x 8x

8x 8x 2 ? 8x 2 ? 23x 2 ? 23x 23x1

from which log28x 8x log223x1 3x 1. Since 8x 8x is positive for every x in R, log28x 8x 3x 1 for every real number. In each case, replace log28x 8x by 3x 1 and solve the resulting equation.

(a) 3x 1 x 1, so x 1; the solution set is 1. (b) 3x 1 3x 1, which is an identity, so the solution set is R. (c) 3x 1 3x, or 0 ? x 1. The solution set is the empty set.

Strategy: The domain of

the log3 function is the set of positive real numbers. For f this requires x 2 5x 6 0, and for g, both x 2 0 and x 3 0.

EXAMPLE 8 Domains of logarithmic functions If f x log3x 2 5x 6 and gx log3x 2 log3x 3, then find the domain of each function. Are functions f and g equal? Explain.

Solution Follow the strategy. To find the domain of function f , solve the inequality

x2 5x 6 0, or x 2x 3 0.

The solution set is x x 2 or x 3, so the domain of f is , 2 3, . For the function g, the strategy emphasizes that both x 2 and x 3. The

solution set is x x 3, so the domain of g is 3, . Finally, since functions f and g have different domains, they cannot be equal.

However, f x gx for all x 3.

EXAMPLE 9 Finding an inverse function If f x lnx 1 2,

(a) Find a formula for f 1. Graph f and f 1 on the same screen. (b) Describe the graphs in terms of transformations of the natural exponential and

logarithmic functions.

Solution

(a) Using the algorithm from Section 2.7, we write y f x, interchange vari-

ables, and solve for y.

y lnx 1 2

x lny 1 2

lny 1 x 2

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