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594 REVIEW OF MATHEMATICS FOR TELECOMMUNICATION APPLICATIONS

probably mean “the power expressed in dBW.” We might discuss the velocities of two cars. The velocity of car 1 may be expressed as V1, and the velocity of car 2 as V2. The use of the subscript allows us to distinguish between the two cars. Subscripts are used widely in the text.

Independent and Dependent Variables. In the selection of symbols to solve a particular problem, we must distinguish between constants and variables specific to the problem at hand. Consider the volume of a circular cone where we keep its height constant (h). The formula for its volume is V pr2h/ 3, where r is the radius of the base circle. The height, h, of course, is constant cfor this problem, and just for this problem alone. It is referred to as a parameter. The radius, r, is the independent variable and V is the dependent variable.

B.2.2 Function Concept

We remember the equation for power: P I2R. Let R be 75 Q ; then P 75I2. When

the value of I is known, we can calculatecthe power. For example, whencI

c

2 A, the

power will be:

 

P c 75 × 2 × 2 c 300 W.

That is to say, that the variable P depends on the variable I.

In general, we can say that if a variable y depends on another variable x so that for each value of x (in a suitable set), a corresponding value of y is determined. The variable y then is called a function of x. In symbols this is written as:

y c f (x).

Read this as “y is the function of x” or just “y equals f of x.” Remember here that f is not a qualitative symbol but an operative symbol.

Take the conversion formula from absolute temperature to centrigrade: T 273 + C.

Here T f(C). Another example is noise power. It is a function of absolute temperaturec

and bandwidthc

. Thus:

 

Pn c −228.6 dBW + 10 log T + 10 log BHz.

This is Eq. (9.12) from Chapter 9 of the text. Note the use of an absolute constant, in this case Boltzmann’s constant. Pn is the independent variable; T, the noise temperature in kelvins, and B, the bandwidth in Hz, are dependent variables. We could set the bandwidth at 1 MHz. Then Pn c f(T).

B.2.3 Using the Sigma Notation

The Greek letter S (capital sigma) indicates summation. The index of summation limits the number of items to be summed or added up. The letter i is often used for this purpose. Values for i are placed above and below sigma. The initial value is placed below, and the final value is placed above the letter sigma. For example, we could have

Swi c w1 + w2 + w3 + w4 . . . + w9 + w10.

Here the initial value is 1 and the final value is 10.

B.3 INTRODUCTORY ALGEBRA

595

B.3 INTRODUCTORY ALGEBRA

B.3.1 Review of the Laws of Signs

If we multiply a +factor by another +factor, the product will have a + sign.

If we divide a +factor by another +factor, the quotient will have a + sign.

 

 

If we multiply a

factor by a +factor, the product will have a

sign (i.e., will be

 

 

 

negative).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Likewise, if we multiply a +factor by a factor, the product will have a

sign.

 

 

If we divide a +factor by a factor, the quotient will have a

sign.

 

sign.

 

 

Likewise, if we divide a

factor by a +factor, the quotient will have a

 

 

If we divide a factor by a factor, the quotient will have a + sign.

 

 

; +/

In other words, + × +

 

+;

+/ +

 

+;

× +

c −

; + ×

− c −

;

 

/ +

 

;

×

− c

+; and

/

c

c

+.

 

c

 

 

 

 

 

 

c

− c

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.3.2 Conventions with Factors and Parentheses

Two symbols placed together imply multiplication. Each symbol or symbol grouping is called a factor. Examples:

xy means x multiplied by y;

x( y + 1) means x multiplied by the quantity ( y + 1);

abc means a multiplied by b that is then multiplied by c.

The use of parentheses is vital in algebra. If a parentheses pair has a sign in front of it, that sign is operative on each term inside the parentheses. If the sign is just a +, then consider that each term inside the parentheses is multiplied by +1. Here we just have a flow through.

Example. Simplify 3X + (7X + Y 10) c 3X + 7X + Y 10 c 10X + Y 10.

 

 

 

 

Suppose there is a minus sign in front of the parentheses. Assume it is a

1. For the

terms inside the parentheses, we change each sign and add.

 

 

 

 

 

 

 

 

 

 

 

Consider nearly the same example: Simplify 3X

(7X + Y

10)

c

3X

 

7X

Y +

10

 

4X Y + 10.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Onec −

more

example: Simplify 4.5K (

27.3 + 2.5K)

4.5K + 27.3

2.5K

2K +

27.3. We carry this one step further byplacing

a factor inc front of the parentheses− c

.1

 

 

Example. Simplify: 4X 5Y(X 2 + 2X Y + 10)

36

/

4X

5X 2Y

10X Y 2

 

50Y

36.

Another example: Simplify:

4Q

21(5Q2

/

 

 

 

4)

 

4Q 105Q2

/

 

+

63Q

/

6

 

33Qc6

c

3

 

+ 84

 

4Q

35Q2 + 21Q/ 2 + 84

29Q/ 2

 

35Q2 + 84.

 

 

 

 

 

 

 

 

 

 

 

 

c

 

 

c

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

An algebraic expression may have parentheses inside brackets. The rule is to clear

the “outside” first. In other words, clear the brackets. Then clear the parentheses in that order.

1A factor is a value, a symbol that we multiply by. To factor an expression is to break the expression up into components and when these components are multiplied together, we derive the original expression. For example 12 and 5 are factors of 60 because 12 × 5 c 60.

596

REVIEW OF MATHEMATICS FOR TELECOMMUNICATION APPLICATIONS

 

 

 

 

 

 

 

 

Example. Simplify: (r

3R)

[(2R

r) + 1]

c

(r

3R)

(2R

r)

1

c

r

3R

2R +

r 1 c 2r 5R 1.

 

 

 

 

 

 

 

 

 

 

 

Fractions. We review the adding and subtracting of fractions in arithmetic. Add 3/ 4

+ 5/ 8 + 5/ 12. Remember that we look for the least common denominator (LCD). In

this case it is 24. The reason: 24 is divisible by 4, 8, and 12. The three terms are now converted to 24ths or:

 

18/

24 + 15/ 24 + 10/ 24.

Now add 18, 15, and 10, which is

43/ 24 or 1 19/ 24.

Another example: Add 1/ 6 + 1/ 7 + 1/ 8. The best we can do here for the LCD is

the product of 6 × 7 × 8

336. Convert each fraction to 336 or 56/ 336 + 48/ 336 +

42/ 336. Now add 56 + 48c

+ 42

146/ 336. By inspection we see we can divide the

numerator and denominator by 2 andc

the answer is 73/ 168.

Subtraction of fractions follows the same procedure, just follow the rules of signs.

Example: Calculate 1/ 4 1/ 5 + 2/ 3.

Procedure. Multiply the denominators together for the LCD, which is 60. Convert each

fraction to 60ths or 15/ 60

12/ 60 + 40/

60. We now have a common denominator so

we can add the numerators:15

12 + 40

c

 

43/ 60.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

We apply the same procedure when we add/ subtract algebraic symbols.

 

 

 

 

 

Example. Simplify 1/ (X

4)

1/ (X

5).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4)]/ (X

Procedure. The LCD is the product (X

 

4)(X

5). We then have [(X

5)

(X

4)(X 5)

1/

(X 2

9X + 20).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Then

multiplyc −

thetwo factors (X

 

 

4) and (X 5). There are two approaches. Do

“long” multiplication as we would dowith arithmetic:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

X

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

× X5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5X + 20

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

X 2

 

4X

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sum

 

X 2

9X + 20.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The second approach: (X

 

4) × (X

 

 

5); multiply the leftmost terms (the Xs) and we

get X 2; multiply the right sides

 

( 4 and

 

 

5) and we get +20; then multiply the means

together (

4X ) and the extremestogether

(

 

 

5X ) and add (

 

4X +

 

5X )

 

 

9X. Place

together indescending order:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c −

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

X 2

 

9X + 20.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

There is a grouping we should recognize by inspection in the generic type of (X 2

K 2),

which factors into (X

K)(X + K). Here are several examples: (X 2

1), which factors

into (X + 1)(X

1) or

X 2

64, which factors into (X + 8)(X

8).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.3 INTRODUCTORY ALGEBRA

597

Adding and Subtracting Exponents. An exponent is a number at the right of and above a symbol. The value assigned to the symbol with this number is called a power of the symbol, although power is sometimes used in the same sense as exponent. If the

exponent is a positive integer and X denotes the symbol, then Xn means X if n 1.

When n > 1 we would have 31

c

3, 32

c

9, 33

c

27, and so on. Note that X0

c

1 cif X

itself is not zero.

 

 

 

 

Rules: When we multiply, we add exponents; when we divide, we subtract exponents.

For the preceding zero example, we can think of it as X 2/

X 2 X0 1. This addition

and subtraction can be carried out so long as there is a common cbase. cIn this case it was

X. For example, 23 × 22

25. Another example: X7/ X5

X 2. Because it is division,

we subtracted exponents; itc had the common base “X.”

c

 

A negative exponent indicates, in addition to the operations indicated by the numerical value of the exponent, that the quantity is to be made a reciprocal.

Example 1:

X2

1/ 16.

Example 2:

X3 c 1/ X3.

 

c

 

Further, when addition is involved, and the numbers have a common exponent, we

can just add the base numbers. For example, 3.1 × 1010 + 1.9 × 1010 5.0 × 1010.

We cannot do this if there is not a common base and exponent. The powerc

of ten is

used widely throughout the text.

 

 

 

 

If the exponent is a simple fraction such as 1 or

1 , then we are dealing with a root

2

3

 

 

of the symbol or base number. For example, 91/ 2

3; or (X 2 + 2X + 1)1/ 2

X + 1.

2

3

. First we square X andc then take

Carry this one step further. Suppose we have xc/

 

the cube root of the result. The generalized case is Xp/ q

(Xp)1/ q. In other words, we

first take the pth power and from that result we take the qthc

root. These calculations are

particularly easy to do with a scientific calculator using the Xy function, where y can even be a decimal such as 3.7.

B.3.3 Simple Linear Algebraic Equations

An equation is a statement of equality between two expressions. Equations are of two types: identities and conditional equations (or usually simply equations). A conditional

equation is true only for certain values of the variables involved, for example, x + 2

 

5 is a true statement only when x 3; and xy + y

3

0 is true when x 2 and y c

1, and for many other pairs of valuesc

of x and y; but

forc

still others it is falsec .

c

Equation (9.20) in Chapter 9 of the text is an identity. It states:

G/ T c G 10 log T,

where G is gain and T is noise temperature. Actually the right-hand side of the equation is just a restatement of the left-hand side; it does not tell us anything new. In many cases, identities such as this one are very useful in analysis.

There are various rules for equations. An equation has a right-hand side and a lefthand side. Maintaining equality is paramount. For example, if we add some value to the left side, we must add the same value to the right side. Likewise, if we divide the (entire) left side by a value, we must divide the (entire) right by the same value. As we might imagine, we must carry out similar procedures for subtraction and multiplication.

A linear equation is of the following form: AX + B c 0 (A 0). This is an equation

598 REVIEW OF MATHEMATICS FOR TELECOMMUNICATION APPLICATIONS

with one unknown, X. A will be a fixed quantity, a number; so will B. However, in the parentheses it states that A may not be 0.

Let us practice with some examples. In each case calculate the value of X.

X + 5 c 7

Clue: We want to have X alone on the left side. To do this we subtract 5 from the left side, but following the rules, we must also subtract 5 from the right side. Thus:

X + 5

5

 

7

 

5

or

X

c

2.

 

c

 

 

 

 

 

 

 

 

 

 

 

Another example is

3X + 7 c 31.

Again, we want X alone. But first we must settle for 3X. Subtract 7 from both sides of the equation.

3X + 7

7

 

31

 

7;

3X

c

24.

 

c

 

 

 

 

 

 

 

 

 

Again, we want X alone. To do this we can divide by 3 (each side).

3X/

3

c

24/ 3;

X

 

 

 

c

8.

 

Still another example:

z 2 + 1 c 65 (solve for z). Subtract 1 from each side to get z 2 alone.

z 2

+ 1

1

c

65

1;

 

2

 

 

 

 

z

 

c 64.

 

 

 

 

 

Take the square root of each side. Thus

z c 8.

More Complex Equations

Solve for R.

0.25(0.54R + 2.45)

0.24(2.3R 1.75)

 

0.135R + 0.6125 c

0.552R

0.42

 

0.552R + 0.135R c

 

0.42

 

0.6125

c

 

 

0.417R c

1.0325

 

 

R c

2.476.

 

 

Another example: Solve for x.

 

 

 

 

 

 

 

 

 

B.3 INTRODUCTORY ALGEBRA

599

(x + 4)(x

3)

c

(x

9)(x

2)

 

 

2

 

 

2

 

 

 

x

 

12 c x

 

11x + 18

 

 

 

12xx c

230.5.

 

 

 

 

 

 

 

 

c

 

 

 

 

 

 

 

B.3.4. Quadratic Equations

Quadratic equations will have one term with a square (e.g., X 2) and they take the form:

Ax 2 + Bx + C

c

0

(A 0)

 

 

 

where A, B, and C are constants (e.g., numbers). A quadratic equation should always be set to 0 before a solution is attempted. For instance, if we have an equation that is

2x 2 + 3x

21, convert this equation to: 2x 2

+ 3x + 21

0.

 

 

We willc discuss

two methods of solving a quadratic equation:c

by factoring and by

the quadratic formula.

 

 

 

 

 

 

 

Factoring. Suppose we have the simple relation: x 2

1

0. We remember from the

preceding that this factors into (x 1)(x + 1)

0. Thisbeingc

the case, at least one of

the factors must equal 0. If this is not

understood,c

realize that there is no other way for

the equation statement to be true. Keep in mind that anything multiplied by 0 will be 0. So there are two solutions to the equation:

x 1 c 0, thus x c 1 or x + 1 c 0 and x c −1

Proof that these are correct answers is by substituting them in the equation. Solve for x in this example:

 

 

 

 

 

x 2

 

100x + 2400

 

0.

 

 

 

 

 

 

 

 

 

 

 

c

 

 

 

 

 

This factors into: (x

40)(x

60)

 

0. We now have two factors: x 40 and x

60,

whose product is 0. This

means

thatc we must have either: x 40 0,where x

40

or

x 60 0 and in this case x

60. We can check our results by substitutionc

thatc either

ofthesec values satisfies the equationc

.

 

 

 

 

12. Multiply the factors: x 2

 

 

Another example: Solve for x in (x 3)(x

2)

 

5x +

6, then x 2 5x + 6

12. Subtract 12 from

both

sidesc

of the equation so that we set the

left-hand side

equalcto 0. Thus:

 

 

 

 

 

 

 

 

 

 

 

x 2 5x 6 c 0, factors into (x 6, )(x + 1). c 0 Then x 6 c 0 x c 6 or x + 1 c 0 x c −1

Quadratic Formula. This formula may be used on the conventional quadratic equation in the generic form of

Ax 2 + Bx + C

c

0

(A 0).

 

 

 

The value of x is solved by simply manipulating the constants A, B, and C. The quadratic formula is stated as follows:

600 REVIEW OF MATHEMATICS FOR TELECOMMUNICATION APPLICATIONS

x c [B ± (B2 4AC)1/ 2]/ 2A or, rewritten with the radical sign:

 

 

Bf

±B2

4AC

 

.

x c

 

2A

 

 

Just as we did with the factoring method, the quadratic formula will produce two roots (two answers): one with the plus before the radical sign and one with the minus before the radical sign.

Example 1. Solve for x: 3x 2

2x

5

c

0. Here A

c

3, B

c −

2 and C

c −

5. Apply the

quadratic formula.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x

c

(+2 ± 4 + 60)/ 6

 

 

 

 

 

 

 

 

 

 

The first possibility is (+2 + 64)/ 6

 

 

 

 

 

 

 

 

 

 

 

8)/ 6

1. The

 

 

10/ 6; the second possibility is (+2

quadratic formula will not handle thec

square root of a negative number.The squarec −

root

of a negative number can usually be factored down to (

1)1/ 2, which, by definition, is

the imaginary number i, and is beyond the scope of thisappendix.

 

 

 

 

 

 

Example 2. Solve for E: E 2

3E

 

2

 

0. A 1, B

 

3 and C

2. Thus E

(+3 ± 9

+ 8)/ 2. The first possibility is(3 +4.123c

)/ 2 andc

thecsecond

possibilityc −

is (3

c4.123)/ 2.

Thus E c 3.562 or 0.562.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.3.5 Solving Two Simultaneous Linear Equations with Two Unknowns

There are two methods of solving two simultaneous equations:

1. The graphical method, where both equations are plotted and the intersection of the line derived is the common solution; and

2. The algebraic solution.

We will concentrate on the algebraic solution. There are two approaches to solving two simultaneous equations by the algebraic solution:

1. Elimination; and

2. Substitution.

Elimination Method. With this method we manipulate one of the equations such that when the two equations are either added or subtracted, one of the unknowns is eliminated. We then solve for the other unknown. The solution is then substituted in one of the original equations and we solve for the other unknown.

Example:

2x + 3y 8 c 0. 4x 5y + 6 c 0

Multiply each term by 2 in the upper equation, and we derive the following new equation:

B.3 INTRODUCTORY ALGEBRA

601

4x + 6y 16 c 0.

Place the second equation directly below this new equation, and subtract.

4x + 6y 16

0

4x

5y+ 6 c

0.

 

c

 

If we subtract the lower equation from the upper, we eliminate the 4x term. Now solve for y.

+11y 1122y c 022

y cc 2.

Substitute y 2 in the original upper equation. Then 2x + 6 8

0, 2x

2 and x 1. So

the solution cof these equations is x 1 and y 2. Check the

csolutionsc

by substitutingc

these values into the two original simultaneousc cequations.

 

 

Example:

3x 2y 5 c 0. 6x + y + 12 c 0

There are several possibilities to eliminate one of the unknowns. This time let us multiply each term in the lower equation by two and we get:

12x + 2y + 24 c 0.

Place this new equation below the original upper equation:

123xx+22yy+245 c

00.

c

 

Add the two equations and we get 15x + 19 c 0. Solve for x.

 

 

 

 

 

15xx c −

1919/ 15.

 

 

 

 

 

 

 

 

 

 

c −

 

 

 

 

 

Substitute this value in the upper equation and solve for y.

 

 

3(

 

19/ 15)

 

2y 5

0

 

 

 

 

 

 

 

57/ 15

75/15 c

2y

 

 

66/ 15 or

 

22/ 5.

 

2y c

132/ 15 and y

c −

 

 

 

 

c

 

 

 

 

Substitution Method. Select one of the two simultaneous equations and solve for one of the unknowns in terms of the other. Example: (repeating the first example from above)

loga y c x.

602 REVIEW OF MATHEMATICS FOR TELECOMMUNICATION APPLICATIONS

2x + 3y 8 c 0. 4x 5y + 6 c 0

We can select either equation. Select the first equation. Then:

2xx c

(8833yy)/ 2.

c

Substitute this value for x in the second equation.

4(8 3y)/ 2 5y + 6 c 0

16 6y 5y 11+ 6y c 0 22

yc 2. c

B.4 LOGARITHMS TO THE BASE 10

B.4.1 Definition of Logarithm

If b is a positive number different from 1, the logarithm of the number y, written loga y, is defined as follows: if ax c y, then x is a logarithm of y to the base a, and we write:

This shows, therefore, that a logarithm is an exponentthe exponent to which the base is raised to yield the number. The expression loga y is read: “logarithm of y to the base

a.” The two equations ax y and loga y

x are two different ways of expressing the

relationship between the numbersc

x, y, andc

a. The first equation is in the exponential

form and the second is in the logarithmic form. Thus 26

64 is equivalent to log2 64

6. Likewise, the statement log16(1/ 4)

1/ 2 implies 16c

1/ 2

1/ 4. These concepts

shouldc

be thoroughly understood before cproceeding.

c

 

B.4.2 Laws of Logarithms

In Section B.3 we discussed the laws of adding and subtracting exponents. From these laws we can derive the laws of logarithms. Let us say that the generalized base of a logarithm is a, which is positive, and that x and y are real numbers. Here we mean they are not imaginary numbers (i.e., based on the square root of 1).

Note that a can be any positive number. However, we concentrate

on a

10, that

is, on logarithms to the base 10. The scientific calculator should be used to

obtainc

the

logarithm by using the “log” button. There will also probably be an “ln” button. This button is used to obtain logarithms to the natural base, where a c 2.71828183+.

Law 1. The logarithm of the product of two numbers equals the sum of the logarithms of the factors. That is:

loga xy c loga x + loga y.

Law 2. The logarithm of the quotient of two numbers equals the logarithm of the dividend minus the logarithm of the divisor. That is:

 

 

 

B.4

LOGARITHMS TO THE BASE 10

603

Table B.1 Selected Powers of Ten

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Logarithm of

 

 

 

 

 

 

Power of 10

Number

Number

Value of Logarithm

 

 

 

 

 

 

 

 

 

 

 

104

 

10,000

log10,000

 

4

 

 

 

 

103

 

1000

log1000

 

3

 

 

 

 

102

 

100

log100

 

2

 

 

 

 

101

 

10

log10

 

1

 

 

 

 

100

 

1

log1

 

0

 

 

 

 

101

0.1

log0.1

1 or 9

10

 

 

10−

2

0.01

log0.01

2 or 8

10

 

 

3

 

 

10

4

0.001

log0.001

3 or 7

10

 

 

10

 

0.0001

log0.0001

4 or 6

10

 

 

 

 

 

 

 

 

 

 

loga x/ y c loga x loga y.

Law 3. The logarithm of the nth power of a number equals n times the logarithm of the number. That is:

loga xn c n loga x.

Law 4. The logarithm of the pth root of a number is equal to the logarithm of the number divided by n. That is:

 

 

log(x)1/ p

1/ p loga x.

 

 

c

 

Remember that if x 1, loga x

0. Here is an exercise. Express log10(38)1/ 2(60)/ (29)3

c

1/ 2 log10(38) + logc10(60) 3 clog10(29).

 

The logarithm of a number

has two components: its characteristic and its mantissa.

The characteristic is an integer and the mantissa is a decimal. If the number in question is 10 or its multiple, the logarithm has a characteristic only, and its mantissa is .000000++. Consider Table B.1 containing selected the powers of 10.

A scientific calculator gives both the characteristic and the mantissa when a number is entered and we press the “log” button. On most calculators, just enter the number then press the “log” button. Its logarithm (to the base 10) will then appear in the display.

Table B.2 gives ten numbers and their equivalent logarithms. One of the logarithms is blatantly in error. Identify it.

Deriving one type of logarithm may be tricky. This group consists of decimals, in other words, numbers less than one (< 1). First check column 4 of Table B.1.

Find the logarithm of 0.00783. In scientific notation we can express the number as

7.83 × 103. Therefore, log 0.00783 log(7.83 × 103) log(7.83) + log(103).

Suppose we are given the logarithmc

of a number, assumingc

the base is 10, and we

wish to find the number that generated that logarithm. This is shown on the calculator, usually on the second level, log1. Sometimes in the literature it is indicated as the antilog.

Example. Given the logarithm, find its corresponding number. Use a scientific calculator.

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