Jacques I. Mathematics for Economics and Business

Подождите немного. Документ загружается.

(1) Use your graphs to estimate the values of Q for which the firm

(a) breaks even (b) maximizes profit

(2) Confirm your answer to part (1) using algebra.

10 The profit function of a firm is of the form

π=aQ

+ bQ + c

If it is known that π=9, 34 and 19 when Q = 1, 2 and 3 respectively, write down a set of three simul-

taneous equations for the three unknowns, a, b and c. Solve this system to find a, b and c. Hence find

the profit when Q = 4.

11 (Excel) Tabulate values of the total cost function

TC = 0.01Q

+ 0.5Q

+ Q + 1000

when Q is 0, 2, 4,..., 30 and hence plot a graph of this function on the range 0 ≤ Q ≤ 30. Use this

graph to estimate the value of Q for which TC = 1400.

12 (Excel) A firm’s total revenue and total cost functions are given by

TR =−0.5Q

+ 24Q

TC = Q√Q + 100

Sketch these graphs on the same diagram on the range 0 ≤ Q ≤ 48. Hence estimate the values of

Q for which the firm

(a) breaks even (b) maximizes profit

Non-linear Equations

140

MFE_C02b.qxd 16/12/2005 10:59 Page 140

section 2.3

Indices and logarithms

Objectives

At the end of this section you should be able to:

 Evaluate b

in the case when n is positive, negative, a whole number or fraction.

 Simplify algebraic expressions using the rules of indices.

 Investigate the returns to scale of a production function.

 Evaluate logarithms in simple cases.

 Use the rules of logarithms to solve equations in which the unknown occurs as a

power.

Advice

This section is quite long with some important ideas. If you are comfortable using the rules

of indices and already know what a logarithm is, you should be able to read through the

material in one sitting, concentrating on the applications. However, if your current under-

standing is hazy (or non-existent), you should consider studying this topic on separate

occasions. To help with this, the material in this section has been split into the following

convenient sub-sections:

 index notation

 rules of indices

 logarithms

 summary

MFE_C02c.qxd 16/12/2005 11:00 Page 141

2.3.1 Index notation

We have already used b

as an abbreviation for b × b. In this section we extend the notation to

for any value of n, positive, negative, whole number or fraction. In general, if

M = b

we say that b

is the exponential form of M to base b. The number n is then referred to as the

index, power or exponent. An obvious way of extending

= b × b

to other positive whole-number powers, n, is to deﬁne

= b × b × b

= b × b × b × b

and, in general,

To include the case of negative powers, consider the following table of values of 2

−3

−2

−1

????24816

To work from left to right along the completed part of the table, all you have to do is to

multiply each number by 2. Equivalently, if you work from right to left, you simply divide

by 2. It makes sense to continue this pattern beyond 2

= 2. Dividing this by 2 gives

= 2 ÷ 2 = 1

and dividing again by 2 gives

−1

= 1 ÷ 2 =

and so on. The completed table is then

−3

−2

−1

/2 124 816

Notice that

−1

−2

−3

In other words, negative powers are evaluated by taking the reciprocal of the corresponding

positive power. Motivated by this particular example, we deﬁne

= b × b × b × ...b

Non-linear Equations

142

MFE_C02c.qxd 16/12/2005 11:00 Page 142

and

where n is any positive whole number.

−n

= 1

2.3 • Indices and logarithms

143

Example

Evaluate

(a) 3

(b) 4

(d) 5

(e) 5

−1

(f) (−2)

(g) 3

−4

(h) (−2)

−3

(i) (1.723)

Solution

Using the deﬁnitions

= b × b × b × ...× b

= 1

−n

we obtain

(a) 3

= 3 × 3 = 9

(b) 4

= 4 × 4 × 4 = 64

= 1

(d) 5

= 5

(e) 5

−1

(f) (−2)

= (−2) × (−2) × (−2) × (−2) × (−2) × (−2) = 64

where the answer is positive because there are an even number of minus signs.

(g) 3

−4

== =

(h) (−2)

−3

== =−

where the answer is negative because there are an odd number of minus signs.

(i) (1.723)

= 1

because any number raised to the power of zero equals 1.

(−2) × (−2) × (−2)

(−2)

3 × 3 × 3 × 3

MFE_C02c.qxd 16/12/2005 11:00 Page 143

We handle fractional powers in two stages. We begin by deﬁning b

where m is a reciprocal

such as

and then consider more general fractions such as

later. Assuming that

n is a positive whole number, we deﬁne

By this we mean that b

1/n

is a number which, when raised to the power n, produces b. In sym-

bols, if c = b

1/n

then c

= b. Using this deﬁnition,

1/2

= square root of 9 = 3 (because 3

= 9)

1/3

= cube root of 8 = 2 (because 2

= 8)

625

1/4

= fourth root of 625 = 5 (because 5

= 625)

Of course, the nth root of a number may not exist. There is no number c satisfying c

=−4,

for example, and so (−4)

1/2

is not deﬁned. It is also possible for some numbers to have more

than one nth root. For example, there are two values of c which satisfy c

= 16, namely c = 2 and

c =−2. In these circumstances it is standard practice to take the positive root, so 16

1/4

= 2.

We now turn our attention to the case of b

, where m is a general fraction of the form p/q

for some whole numbers p and q. What interpretation are we going to put on a number such

as 16

3/4

? To be consistent with our previous deﬁnitions, the numerator, 3, can be thought of as

an instruction for us to raise 16 to the power of 3, and the denominator tells us to take the

fourth root. In fact, it is immaterial in which order these two operations are carried out. If we

begin by cubing 16 we get

= 16 × 16 × 16 = 4096

and taking the fourth root of this gives

3/4

= (4096)

1/4

= 8 (because 8

= 4096)

On the other hand, taking the fourth root ﬁrst gives

1/4

= 2 (because 2

= 16)

and cubing this gives

3/4

= 2

= 8

which is the same answer as before. We therefore see that

(16

)

1/4

= (16

1/4

)

This result holds for any base b and fraction p/q (provided that q is positive), so we deﬁne

p/q

= (b

)

1/q

= (b

1/q

)

1/n

= nth root of b

Non-linear Equations

144

Practice Problem

1 (1) Without using a calculator evaluate

(a)

(b) 10

(d) 10

−1

(e) 10

−2

(f) (−1)

100

(g) (−1)

(h) 7

−3

(i) (−9)

(j) (72 101)

(k) (2.718)

(2) Confirm your answer to part (1) using a calculator.

MFE_C02c.qxd 16/12/2005 11:00 Page 144

2.3.2 Rules of indices

There are two reasons why the exponential form is useful. Firstly, it is a convenient shorthand

for what otherwise might be a very lengthy number. The exponential form

is much easier to write down than either of the equivalent forms

9 × 9 × 9 × 9 × 9 × 9 × 9 × 9

43 046 721

2.3 • Indices and logarithms

145

Advice

Given that we are allowed to perform these operations in any order, it is usually easier to

find the qth root first to avoid having to spot roots of large numbers.

Example

Evaluate

(a) 8

4/3

(b) 25

−3/2

Solution

(a) To evaluate 8

4/3

we need both to raise the number to the power of 4 and to ﬁnd a cube root. Choosing

to ﬁnd the cube root ﬁrst,

4/3

= (8

1/3

)

= 2

= 16

(b) Again it is easy to ﬁnd the square root of 25 ﬁrst before raising the number to the power of −3, so

−3/2

= (25

1/2

)

−3

= 5

−3

For this particular exponential form we have actually carried out three distinct operations. The minus

sign tells us to reciprocate, the fraction

/2 tells us to take the square root and the 3 tells us to cube. You

might like to check for yourself that you get the same answer irrespective of the order in which these

three operations are performed.

125

Practice Problem

2 (1) Without using your calculator, evaluate

(a) 16

1/2

(b) 27

1/3

5/2

(d) 8

−2/3

(e) 1

−17/25

(2) Confirm your answer to part (1) using a calculator.

MFE_C02c.qxd 16/12/2005 11:00 Page 145

Non-linear Equations

146

Secondly, there are four basic rules of indices which facilitate the manipulation of such

numbers. The four rules may be stated as follows:

Rule 1 b

××

Rule 2 b

/ b

−−

Rule 3 (b

)

Rule 4 (ab)

It is certainly not our intention to provide mathematical proofs in this book. However, it might

help you to remember these rules if we give you a justiﬁcation based on some simple examples.

We consider each rule in turn.

Rule 1

Suppose we want to multiply together 2

and 2

. Now 2

= 2 × 2 and 2

= 2 × 2 × 2 × 2 × 2, so

× 2

= (2 × 2) × (2 × 2 × 2 × 2 × 2)

Notice that we are multiplying together a total of seven 2s and so by deﬁnition this is just 2

that is,

× 2

= 2

2+5

This conﬁrms rule 1, which tells you that if you multiply two numbers, all you have to do is to

add the indices.

Rule 2

Suppose we want to divide 2

by 2

. This gives

Now, by deﬁnition, reciprocals are denoted by negative indices, so this is just 2

−3

: that is,

÷ 2

= 2

−3

= 2

2−5

This conﬁrms rule 2, which tells you that if you divide two numbers, all you have to do is to

subtract the indices.

Rule 3

Suppose we want to raise 10

to the power 3. By deﬁnition, for any number b,

= b × b × b

so replacing b by 10

we have

(10

)

= 10

× 10

= (10 × 10) × (10 × 10) × (10 × 10) = 10

because there are six 10s multiplied together: that is,

(10

)

= 10

2×3

This conﬁrms rule 3, which tells you that if you take a ‘power of a power’, all you have to do is

to multiply the indices.

2 × 2 × 2

2 × 2

2 × 2 × 2 × 2 × 2

MFE_C02c.qxd 16/12/2005 11:00 Page 146

Rule 4

Suppose we want to raise 2 × 3 to the power 4. By deﬁnition,

= b × b × b × b

so replacing b by 2 × 3 gives

(2 × 3)

= (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3)

and, because it does not matter in which order numbers are multiplied, this can be written as

(2 × 2 × 2 × 2) × (3 × 3 × 3 × 3)

that is,

(2 × 3)

= 2

× 3

This conﬁrms rule 4, which tells you that if you take the power of a product of two numbers,

all you have to do is to take the power of each number separately and multiply.

A word of warning is in order regarding these laws. Notice that in rules 1 and 2 the bases of

the numbers involved are the same. These rules do not apply if the bases are different. For

example, rule 1 gives no information about

× 3

Similarly, please notice that in rule 4 the numbers a and b are multiplied together. For some

strange reason, some business and economics students seem to think that rule 4 also applies to

addition, so that

(a + b)

= a

+ b

This statement is NOT TRUE.

It would make algebraic manipulation a whole lot easier if it were true, but I am afraid to say

that it is deﬁnitely false! If you need convincing of this, note, for example, that

(1 + 2)

= 3

= 27

which is not the same as

+ 2

= 1 + 8 = 9

One variation of rule 4 which is true is

= (b ≠ 0)

This is all right because division (unlike addition or subtraction) is the same sort of operation

as multiplication. In fact,

can be thought of as

a ×

so applying rule 4 to this product gives

as required.

2.3 • Indices and logarithms

147

MFE_C02c.qxd 16/12/2005 11:00 Page 147

The following example demonstrates how rules 1–4 are used to simplify algebraic

expressions.

Non-linear Equations

148

Advice

There might be occasions (such as in examinations!) when you only half remember a rule

or perhaps think that you have discovered a brand new rule for yourself. If you are ever

worried about whether some rule is legal or not, you should always check it out by trying

numbers, just as we did for (a + b)

. Obviously, one numerical example which actually

works does not prove that the rule will always work. However, one example which fails is

good enough to tell you that your supposed rule is rubbish.

Example

Simplify

(a) x

1/4

× x

3/4

(b) (c) (x

−1/3

)

Solution

(a) The expression

1/4

× x

3/4

represents the product of two numbers in exponential form with the same base. From rule 1 we may

add the indices to get

1/4

× x

3/4

= x

1/4+3/4

= x

which is just x.

(b) The expression

is more complicated than that in part (a) since it involves numbers in exponential form with two dif-

ferent bases, x and y. From rule 2,

may be simpliﬁed by subtracting indices to get

÷ x

= x

2−4

= x

−2

Similarly,

= y

÷ y

= y

3−1

= y

Hence

= x

−2

MFE_C02c.qxd 16/12/2005 11:00 Page 148

It is not possible to simplify this any further, because x

−2

and y

have different bases. However, if you

prefer, this can be written as

because negative powers denote reciprocals.

−1/3

)

is to apply rule 4, treating x

as the value of a and y

−1/3

as b to get

−1/3

)

= (x

)

( y

−1/3

)

Rule 3 then allows us to write

)

= x

2×3

= x

( y

−1/3

)

= y

(−1/3)×3

= y

−1

Hence

−1/3

)

= x

−1

As in part (b), if you think it looks neater, you can write this as

because negative powers denote reciprocals.

2.3 • Indices and logarithms

149

Practice Problem

3 Simplify

(a)

3/4

)

(b) (c) (x

)

(d) √x(x

5/2

+ y

)

[Hint: in part (d) note that √x = x

1/2

and multiply out the brackets.]

3/2

There are occasions throughout this book when we use the rules of indices and deﬁnitions

of b

. For the moment, we concentrate on one speciﬁc application where we see these ideas

in action. The output, Q, of any production process depends on a variety of inputs, known as

factors of production. These comprise land, capital, labour and enterprise. For simplicity

we restrict our attention to capital and labour. Capital, K, denotes all man-made aids to pro-

duction such as buildings, tools and plant machinery. Labour, L, denotes all paid work in the

production process. The dependence of Q on K and L may be written

Q = f(K, L)

which is called a production function. Once this relationship is made explicit, in the form of a

formula, it is straightforward to calculate the level of production from any given combination

of inputs. For example, if

Q = 100K

1/3

1/2

MFE_C02c.qxd 16/12/2005 11:00 Page 149