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THE SLIDE RULE

Within 20 years of the publication of Henry Briggs’ tables of logarithms, the use of logarithms had spread all the way around the world. Logarithms were no longer a strange mathematical technique used only by great scientists like Johannes Kepler: they had become a common tool in schoolrooms. Logarithms were also commonly used in all trades and professions that needed to make frequent calculations. It is hard to imagine an invention which has helped the process of computation more dramatically, the only exceptions being the modern digital calculator and the computer.

At about this time, Briggs’ colleague Edmund Gunter realized that instead of carrying around a set of logarithm tables, he could inscribe distances on a piece of wood to represent the logarithms. A pair of dividers could be used to measure these lengths, and adding the lengths together would be equivalent to adding the logarithms. This device (known as Gunter’s Line of Numbers) quickly became a popular calculating tool.

Of course, it did not take long for William Oughtred to see that in order to measure these distances with a pair of dividers, it is only necessary to have two scales which can slide past one another. This was the basic principle behind a calculating tool called the slide rule. In 1850, a French army officer called Amedee Mannheim took Oughtred’s idea and invented the slide rule that we know today.

A slide rule is a computing tool which uses the properties of logarithms to allow the user to perform complicated calculations easily. Since log(ab) = log a + log b, it is quite easy to understand how a slide rule

works. An advertisement from the days of the slide rule

A slide rule consists of three main parts, as shown in the picture:

The body and slide of the rule are printed with different scales, and each scale has a specific function. The letters to the left and right of the individual scales are the names of the scales: A & B, CI & C, D, K, etc. In its most basic form, the slide rule

uses two logarithmic scales which allow the user to rapidly multiply and divide large numbers. More complex slide rules allow the user to calculate other things such as square roots, exponential expressions, logarithms and trigonometric functions.

To make a calculation on a slide rule, we use one hand to hold the body of the rule and the other to move the slide and cursor to align the different scales. The numbers which are brought into line show the approximate value of the product, quotient or other desired result.

For example, let us begin with a simple product: multiply 2 by 4. We move the left index of the C scale over 2, then move the cursor to 4 on the C scale and finally read the answer 8 on the D scale. Physically, we have added the logarithm of 2 to the logarithm of 4, but we have in fact multiplied the the two numbers.

Notice that on a slide rule, multiplying 2 by 4 is functionally equivalent to multiplying 2 by 400, or 200 by 4 000 000. It is up to you to work out the powers of 10 in each case.

Exponential and Logarithmic Functions

219

For this reason, it is best to express the numbers in scientific notation before using the slide rule. For example, we can write 200 times 4 000 000 as 2.0 × 102 × 4.0 × 106, which reduces to 2.0 × 4.0 × 108, and finally 8 × 108.

Now let us try a harder calculation: What is 2.1 × 5.5 × 7.8 / 32?

To divide two numbers, we subtract the logarithm of the second number from the first. For example, to divide 5.5 by 2, we line up 2 on the C scale with 5.5 on the D scale, then read the answer 2.75 on the left index of the C scale. Although we have physically subtracted two quantities, we have in fact divided one number by another.

Move the cursor to 2.1 D.

Line up 3.2 C with the cursor.

Move the cursor to 5.5 C.

Line up 10 C with the cursor. (32 = 3.2 10)

Move the cursor to 7.8 C.

Read the answer on the D scale: 2.815.

Notice that we cannot use a slide rule for addition or subtraction. We can also calculate square and cube roots on a slide rule, although they require special scales. You can find out more about these scales and about slide rules in general on the Internet.

The pictures below show some different and unusual types of slide rule.

 

 

 

 

 

 

 

 

A lighter with a

 

 

 

 

 

 

 

 

A watch with a powerful two-scale

 

working circular

 

 

 

 

 

 

 

 

 

 

220

 

A circular slide rule

 

slide rule

 

slide rule

 

11

 

 

 

 

 

 

 

 

 

 

 

EXERCISES 3.3

A.Exponential Equations and Inequalities

In questions 1-9, solve the equations for x.

1. a.

4x = 8

b. 23x – 1 =

1

 

 

64

 

 

 

 

 

 

 

 

 

c.

5

x2

x

= 25

2x –1

=

3

x

 

 

d. 3

 

 

e.

23x 5x = 1600

f. 62x+4 = 33x 2x+8

 

 

x

1

 

2x–1

2x–1

g.

74+x – 1 =1

h. 27 x = 9 2

2. a.

2x – 1 = 5

b.

3x2 = 1

c.

5x2 = 4

d. 32x – 1 = 2

e.

23x + 2 = 3

f.

32x – 4 = 0

3. a.

2x 3x+1 5x – 2 = 7

b. 24x 22x 5x = 1

c.

53x+1 = 82x – 1

d. 3x 5x = 72x

e.53(x+2) = 9x+2

4.a. 52x – (6 5x) = 475

b.32x+5 = 3x+2 + 2

c.42x – 2 – (6 4x – 2) – 1 = 0

d.9x – 3x + 2 + 14 = 0

e. (1 + ñ3)x + 2x – 1 (2 + ñ3)x = 4

5.a. e2x e4x =1

b.3x+1 + 3x = 324

c.(2 5x+2) + 5x–1 = 251

d.3x 5x = 22x+1 7x+2

6. a. (4 22x) – 6x – (18 32x) = 0

1

1

1

b.(6 9x )+(6 4x )=13 6 x

c.(7 4x2 )+(2 49x2 )= 9 14x2

d.25x – (3 10x) + (2 4x) = 0

e. (5 50x) + 8x = (2 125x) + (4 20x)

7. a. 2x +1=( 21)2x +( 21)x

b. 2x – 3x = 6x – 9x

c. 32x + 3362x – (3x + 36x ) – 8 = 0

8. a. 2x + 3x = 35

b. 3 + 5x/3 = 2x

9. a. xñx = òxx

b. xx2 –3x+2 =1

In questions 10–13, solve the inequalities.

 

10. a.

5x > 3125

b. 23x – 1 = (

1

)x >

1

 

8

 

 

64

 

c.

5x – 3 73 – x

d. 152x+4 – 33x 54x – 4 0

e. 3x + 2x – 1 + 2x+2 – 3x – 1 + 2x – 3 0

 

11. a.

 

 

3x

 

> 3

 

x

– 2

x

3

 

 

 

b.2x + 3x + 4x 29

c.40x – 9x < 16x + 15x

Exponential and Logarithmic Functions

221

12.a. 0.1x+1 < 0.8 + 2 10x

b.25x + 5x+1 50

c.(3 16x) + (2 81x) – (5 36x) < 0

13. a.

6

 

< 2x

b.

120

 

 

 

1

2x – 1

1– 25x

5

x – 2

 

 

 

 

14.(4x2 +2x + 1)x2 x <1

B. Logarithmic Equations and Inequalities

15.Solve the equations.

a.log2(3x – 4) = 3

b.log3 x + log3(x + 6) = 3

c.log(x + 2) – log(x – 2) = 0

d.log9 x2 – log9(2x + 5) – log9(2x – 5) = – 21

e.log7(2x2 – 5x) = 1

f.log2(log4 x) = 2

g.log4(log2(log 3(2 x– 1)))= 21

16.Solve the equations.

a.log 1 (x2 +3x – 4)= log 1(2x+2)

5

5

b.log3 2x2 – 54 = log3(x – 4) x+3

c.log4(3x + 1) = log4(–2x + 14)

d.log5x – 2(2x2) = log5x –2(x + 1)

17.Solve the equations.

a.eln (2x – 1) – 10log x = 1

b.3log32 x+ log 3 x2 = 5

c.log2 (2x – 3) + x = 2

d.x + log3 (3x – 8) = 2

e.xln x = x

f.3ln x + xln 3 = 54

g. logx[2(1 – x)] – log1 – x(2x) = 0 h. log3(2x + 1) = log2(3x – 1)

18.Solve the equations.

a.log2 x + log4 x + log16 x = 7

b.logx 4 + logx2 64 = 5

c.(3 logx 16) – (4 log16 x) = 2 log2 x

d.xlog3 x +3log 32 x =162

19.Find the solution set for each inequality, as an interval.

a. log4(2x – 5) < 1

b.

ln(x + 2) 0

c. log3/4(2x – 4) > 2

d.

log4(x2 – 9) 2

e. log 3 (x2 – 3)> 0

f.

ln x 0

5

 

 

g.log0.1(x2 + x + 1) – log0.1 3 > 0

h.log3(x – 1) – log3 5 < 1

i.log2(log (x – 3)) 0

j.log 1 (log5( x – 6)) 0

2

k.log2(x – 1) + log2(x – 3) < 3

20.Solve 1+ log2a x >1 for 0 < a < 1. 1+ loga x

222

Algebra 11

C.Systems of Equations and Inequalities

21.Find the ordered pair(s) (x, y) that satisfy each system.

a.

53x

= 54y+7

 

 

b.

4(x y)2 – 1 = 1

 

2x

4y = 16

 

5x+y = 125

 

 

 

 

 

c.

4x+y = 128

 

 

d.

3975x2 y2 –4 =1

53x – 2y – 3 = 1

2x+y+1 = 1

 

 

 

 

 

 

 

 

 

 

 

8

 

e.

22x

– 2x – 2 = 0

f.

3x 5y = 45

 

9xy – 1 = 0

 

 

3y 5x = 75

 

 

 

 

 

 

22. Solve the systems of equations.

 

 

272x

+ 125

y

/3 = 8

 

xy

 

a.

 

b. 3 2 +(3x . 3y )=12

 

312x – 52y = –16

 

3x +3y =10

 

c.

(3 2x)+(2 3y) = 2.75

 

 

2x – 3y = –0.75

 

 

 

 

 

 

 

d.

3x

+ (3 5x+y) = 378

 

 

 

(5 3x) + 5x+y+1 = 640

 

 

 

 

 

e.

5x+2y

+ 9x+2y = 14

 

 

 

32x y + 52x y = 8

 

 

 

 

 

 

 

f.

5x+2y+2 = 7x+2y

 

 

 

3x – 2y+98 = 5x – 2y

 

 

 

 

 

 

 

23. Solve the systems of equations.

 

a.

log2 x+2 log2 y=3

b.

3x 2y = 972

 

x2 + y4 = 16

logñ3(x y) = 2

 

 

c.

3x 2y = 576

d.

x + y = 5

 

log

(y x) = 4

log2 x + log2

y = 2

 

 

 

 

ñ2

 

 

 

 

 

 

e.

x2 + y2 = 90

 

 

 

log3

x + log3 y = 3

 

 

 

 

 

 

 

24. Solve the systems of equations.

a.

log4 x – log x y=

7

b.

y – log3 x = 1

 

6

x4 = 312

 

x y =16

 

 

 

 

 

 

c.

logy x+ log x y= 5

d.

logy x + logx y = 2

 

2

x2 + y = 30

 

x+ y = 20

 

 

 

 

 

 

e. 2(logy x + logx y) = 5 x y = 8

25. Solve each system of inequalities.

a. (

2

x

(

8

x

 

27

3)

 

9)

 

>

64

2x2 –6x–3

<16

 

25x – (6 5x) < 5

b.

log2x – 3 9 < 2

6x+1 +3x+1 > 3x+2 – 6x +3x

c.16 4x4 –5 > 2 x+1

d.

3x+1 + 3x+2 + 3x+3 > 5x+2 + (3 5x+1) – 5x 22 + 2x 0

e.

32x – 2 + 27 12 3x – 1

(3 + 2ñ2)x + (3 – 2ñ2)x > 34

Exponential and Logarithmic Functions

223

A. DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

In this section, we will study the derivative formulas for exponential and logarithmic function. Let us begin with the derivative formulas for exponential functions.

1. Exponential Functions

DERIVATIVE OF NATURAL EXPONENTIAL FUNCTION

(ex) = ex

Proof (Derivative of Natural Exponential Function)

Let f(x) = ex, then by the definition of derivatives

f (x)= lim

f (x+h) f (x)

= lim

ex h ex

= lim

ex (eh 1)

= e

x

lim

eh 1

.

h

 

h

h

 

h

h 0

h 0

h 0

 

 

h 0

 

We are unable to evaluate this limit using the techniques we have learned before. But the calculations in the following table helps us to guess it correctly.

 

 

 

 

 

 

 

 

 

 

h

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h

 

– 0.1

 

 

– 0.01

– 0.001

0

0.001

0.01

0.1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

eh 1

0.9516

0.9950

0.9995

1

1.0005

1.0050

1.0517

 

h

 

 

 

 

 

 

 

 

 

 

 

 

 

The table tells us that,

lim

eh 1

=1.

 

 

 

 

 

 

h

 

 

 

 

 

 

 

 

 

 

h 0

 

 

 

 

 

 

 

 

 

So, f (x) = ex 1 = ex.

EXAMPLE 106 Find the derivative of the function f(x) = x2ex.

Solution By the Product Rule,

f (x)=(x2ex ) =(x2 ) ex + x2(ex ) = 2xex +x2ex = xex(x+2).

224

Algebra 11

 

 

5

 

EXAMPLE

107 Find the derivative of the function f(x) = (ex + 1)

 

.

2

 

Solution By the General Power Rule,

 

 

 

3

3

 

 

 

f (x)= 5(ex 1)2 (ex +1) =

5 ex(ex +1)2.

2

2

 

 

CHAIN RULE FOR NATURAL EXPONENTIAL FUNCTION

 

 

 

(e f(x))

= e f(x) f (x)

 

 

 

 

 

 

108 Find the derivative of the function f(x) = e3x+1.

 

EXAMPLE

 

 

 

Solution By the Chain Rule,

 

 

 

 

 

 

 

f (x)= e3x 1 (3x+1) = 3e3x 1.

 

 

 

 

 

109 Find dxdy if f(x)= xex2 1.

 

 

 

 

 

EXAMPLE

 

 

 

 

 

 

 

 

 

 

 

Solution By the Product Rule,

 

 

 

 

 

 

 

dy =1 ex2 1 + x ex2 1 (x2

1) = ex2 1

+2x2ex2 1 = ex2 1(2x2 +1).

 

 

dx

 

 

 

 

 

 

 

 

 

 

DERIVATIVE OF EXPONENTIAL FUNCTION

 

 

 

 

 

 

 

 

 

 

 

x

 

x

ln a , a

 

 

 

 

(a ) = a

 

 

Proof (Derivative of Exponential Function)

By using the identity a = eln a we can rewrite the expression as

f(x) = ax = eln ax = ex ln a

(by the property of logarithm)

f (x)= exln a ln a = eln ax

ln a

f (x)= ax ln a.

 

EXAMPLE 110 Given that f(x) = 3x, find f (x).

Solution f (x) = 3x ln 3

Exponential and Logarithmic Functions

225

EXAMPLE 111 Find the derivative of the function f(x) = ex 2x.

Solution By the Product Rule,

f (x) = (ex) 2x + ex (2x) = ex2x + ex2xln 2 = (2e)x(ln 2 + 1).

 

 

CHAIN RULE FOR EXPONENTIAL FUNCTION

 

 

 

 

 

 

 

 

 

 

 

 

 

(a

f(x)

) = a

f(x)

ln a f (x) ,

a

 

 

 

 

 

 

 

 

 

 

 

 

 

112 Find the derivative of the function

f(x) = 5x2+1.

 

 

EXAMPLE

 

 

 

 

 

 

 

 

Solution By the Chain Rule,

 

 

 

 

 

 

 

 

 

f (x) 5x2 +1 ln 5 (x2 +1) =(ln 5) 2x 5x2 +1.

 

 

 

113 If f (x)=

32x+1

 

EXAMPLE

x2 +1

, find f (x).

 

 

 

(32x 1 ) (x2

+1) – 32x+1 (x2 +1)

Solution f (x)=

 

 

 

 

 

 

 

 

 

(x2 +1)2

 

 

 

 

 

 

f (x)=

32x 1 ln 3 2 (x2 +1) 2x 32x+1

 

 

 

 

(x2 +1)2

 

 

 

 

 

 

f (x)=

2 32x 1[(x2

+1) ln 3 x]

 

 

 

 

(x2 +1)2

 

 

 

 

(by the Quotient Rule)

(by the Chain Rule)

Check Yourself 38

Find the derivative of each function.

 

 

1.

f(x) = 2xex

2.

g(x) = ex3 + 3x2 + 1

 

3. f(x) = ex3 ex

4.

f(x) = e2x – 1

5.

g(x) = (x + 1)ex2 – 1

6. f(x) = ex 3x

7.

g(x) = 3x2 + x + 1

8.

h(x)=

34x+1

 

 

 

x2 – 1

 

 

 

 

 

 

 

 

Answers

 

 

 

 

 

 

1. 2ex(x + 1)

2. ex3 + 3x2 + 1 (3x2 + 6x)

3. ex3 – ex (3x2 ex)

4. 2e2x – 1

5. ex2 – 1 (2x2 + 2x + 1)

6. ex 3x(1 + ln 3)

7. 3x2 + x + 1 ln 3 (2x + 1)

8.

2 34x 1 (2 ln3 (x2

1) x)

 

 

(x2 1)2

 

 

 

 

 

 

 

 

226

Algebra 11

2. Logarithmic Functions

DERIVATIVE OF NATURAL LOGARITHMIC FUNCTION

 

 

 

 

(ln x) = 1

, x> 0

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

114 Find the derivative of the function f(x) = x ln x.

EXAMPLE

 

Solution By the Product Rule,

 

 

 

 

 

 

 

 

f (x)=(x) ln x+ x (ln x) =1

ln x+ x 1 = ln x+1.

 

 

 

 

 

 

 

x

 

 

115 Find the derivative of the function

f (x)= ln 5 x3 .

EXAMPLE

 

 

 

 

 

 

3

 

 

 

 

 

 

Solution We can write 5 x3 = x5 . Now,

 

 

 

 

 

 

3

3

 

 

3

 

 

 

 

f (x)=(ln x5 ) =( 3 ln x) =

1 =

 

.

 

 

 

5

 

 

 

5

x

 

5x

 

 

116 Find the derivative of the function f(x) = (ln x + x)2.

EXAMPLE

 

Solution Applying the General Power Rule, f (x)= 2(ln x+ x) (ln x+ x)

f(x)= 2(ln x+ x) 1x+1 .

(by the General Power Rule)

(by the Sum Rule and the derivative of ln (x))

CHAIN RULE FOR NATURAL LOGARITHMIC FUNCTION

(ln f(x)) =

f (x)

f (x)

 

 

 

 

 

EXAMPLE 117 Find the derivative of the function f(x) = ln(x2 + 3x + 1).

Solution f (x)=

(x2 +3x+1)

=

2x+3

.

 

 

 

x2 +3x+1

x2 +3x+1

Exponential and Logarithmic Functions

227

DERIVATIVE OF LOGARITHMIC FUNCTION

(loga x) =

 

1

,

x 0,

a 0,

a 1

x

ln a

 

 

 

 

 

 

 

 

 

 

 

 

Proof

EXAMPLE 118

(Derivative of Logarithmic Function) Let f(x) = loga x, then af(x) = x.

Differentiating both sides of the equation af(x) = x with respect to x, we get af(x)(ln a) f (x) = 1.

Note that we are looking for f(x).

1

1

f (x)=

 

=

 

af (x) ln a

xln a

Find the derivative of the function f(x) = log2 x.

 

 

1

 

 

 

 

 

 

 

Solution f (x)= xln 2

 

 

 

 

 

CHAIN RULE FOR LOGARITHMIC FUNCTION

 

 

 

 

 

 

 

 

(loga f(x)) =

f (x)

 

 

 

f (x) ln a

 

 

 

 

 

 

 

 

 

 

 

 

119

Find the derivative of the function f(x) = log (x3

+ 2x).

EXAMPLE

 

 

 

 

2

 

Solution

f (x)=

(x3 +2x)

=

3x2 +2

 

(x3 +2x) ln 2

(x3 +2x)ln 2

 

 

 

120 Find the derivative of the function f(x) = ex2+3x

 

EXAMPLE

log3(2x – 4).

 

Solution By the Product Rule,

 

f (x) = (ex2+3x) log

(2x – 4)

+ ex2+3x (log (2x

3

 

3

f (x)= ex2+3x (2x + 3) log3(2x – 4) + ex2+3x

– 4))

2 . (2x – 4)ln3

228

Algebra 11