# Chapter 31 logarithms

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Logarithms

31Contents:

A Logarithms in base a [3.10]

B The logarithmic function [3.10]

C Rules for logarithms [3.10]

D Logarithms in base 10 [3.10]

E Exponential and logarithmic

equations [3.10]

In Chapter 28 we answered problems like the one above by graphing the exponential function and using

technology to find when the investment is worth a particular amount.

However, we can also solve these problems without a graph using logarithms.

We have seen previously that y = x2 and y =px are inverse functions.

For example, 52 = 25 andp25 = 5.

If y = ax then we say x is the logarithm of y in base a, and write this as x = loga y.

LOGARITHMS IN BASE a [3.10]A

Opening problem#endboxedheading

Tony invests $8500 for n years at 7:8% p.a. compounding annually. The interest rate is fixed for the

duration of the investment. The value of the investment after n years is given by V = 8500 (1:078)ndollars.

Things to think about:

a How long will it take for Tonys investment to amount to $12 000?

b How long will it take for his investment to double in value?

Logarithms were created to be the .inverse of exponential functions

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Y:\HAESE\IGCSE01\IG01_31\625IGCSE01_31.CDR Tuesday, 18 November 2008 11:10:27 AM PETER

For example, since 8 = 23 we can write 3 = log2 8. The two statements 2 to the power 3 equals 8

and the logarithm of 8 in base 2 equals 3 are equivalent, and we write:

23 = 8 , log2 8 = 3Further examples are: 103 = 1000 , log10 1000 = 3

32 = 9 , log3 9 = 24

1

2 = 2 , log4 2 = 12

The symbol ,is equivalent to

In general, y = ax and x = loga y are equivalent statements

and we write y = ax , x = loga y.

Example 1 Self Tutor

Write an equivalent:

a logarithmic statement for 25 = 32 b exponential statement for log4 64 = 3:

a 25 = 32 is equivalent to log2 32 = 5.

So, 25 = 32 , log2 32 = 5.b log4 64 = 3 is equivalent to 4

3 = 64.

So, log4 64 = 3 , 43 = 64.

Example 2 Self Tutor

Find the value of log3 81:

) 3x = 81

) 3x = 34

) x = 4

) log3 81 = 4

EXERCISE 31A

1 Write an equivalent logarithmic statement for:

a 22 = 4 b 42 = 16 c 32 = 9 d 53 = 125

e 104 = 10000 f 71 = 17 g 33 = 127 h 27

1

3 = 3

i 52 = 125 j 2 1

2 = 1p2

k 4p2 = 22:5 l 0:001 = 103

2 Write an equivalent exponential statement for:

a log2 8 = 3 b log2 1 = 0 c log212

= 1 d log2

p2 = 12

e log2

1p2

= 12 f logp2 2 = 2 g logp3 9 = 4 h log9 3 = 12

3 Without using a calculator, find the value of:

a log10 100 b log2 8 c log3 3 d log4 1

e log5 125 f log5(0:2) g log10 0:001 h log2 128

i log212

j log3

19

k log2(

p2) l log2

p8

The logarithm of in baseis the exponent or powerof which gives .

813

3 81

Let log3 81 = x

626 Logarithms (Chapter 31)

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Y:\HAESE\IGCSE01\IG01_31\626IGCSE01_31.CDR Monday, 27 October 2008 3:01:48 PM PETER

The logarithmic function is f(x) = loga x where a > 0, a 6= 1.

Consider f(x) = log2 x which has graph y = log2 x.

Since y = log2 x , x = 2y, we can obtain the table of values:

y 3 2 1 0 1 2 3x 18

14

12 1 2 4 8

Notice that:

the graph of y = log2 x is asymptotic to the y-axis the domain of y = log2 x is fx j x > 0g the range of y = log2 x is fy j y 2 R g

THE INVERSE FUNCTION OF f(x) = loga x

Given the function y = loga x, the inverse is x = loga y finterchanging x and yg) y = ax

So, f(x) = loga x , f1(x) = ax

THE LOGARITHMIC FUNCTION [3.10]B

x

8642

y

O

y xlogx

627Logarithms (Chapter 31)

m log7

3p7

n log2(4p2) o logp2 2 p log2

1

4p2

q log10(0:01) r log

p2 4 s log

p3

13

t log3

1

9p3

4 Rewrite as logarithmic equations:

a y = 4x b y = 9x c y = ax d y = (p3)x

e y = 2x+1 f y = 32n g y = 2x h y = 2 3a

5 Rewrite as exponential equations:

a y = log2 x b y = log3 x c y = loga x d y = logb n

e y = logm b f T = log5a2

g M = 12 log3 p h G = 5 logbm

i P = logpbn

6 Rewrite the following, making x the subject:

a y = log7 x b y = 3x c y = (0:5)x d z = 5x

e t = log2 x f y = 23x g y = 5

x2 h w = log3(2x)

i z = 12 3x j y = 15 4x k D = 110 2x l G = 3x+1

7 Explain why, for all a > 0, a 6= 1: a loga 1 = 0 b loga a = 1

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Y:\HAESE\IGCSE01\IG01_31\627IGCSE01_31.CDR Tuesday, 18 November 2008 11:12:17 AM PETER

Example 3 Self Tutor

Find the inverse function f1(x) for: a f(x) = 5x b f(x) = 2 log3 x

a y = 5x has inverse function x = 5y

) y = log5 x

So, f1(x) = log5 x

b y = 2 log3 x has inverse function

x = 2 log3 y

)x

2= log3 y

) y = 3x2

So, f1(x) = 3x2

EXERCISE 31B

1 Find the inverse function f1(x) for:

a f(x) = 4x b f(x) = 10x c f(x) = 3x d f(x) = 2 3xe f(x) = log7 x f f(x) =

12(5

x) g f(x) = 3 log2 x h f(x) = 5 log3 x

i f(x) = logp2 x

2 a On the same set of axes graph y = 3x and y = log3 x.

b State the domain and range of y = 3x.

c State the domain and range of y = log3 x.

3 Prove using algebra that if f(x) = ax then f1(x) = loga x.

x

y

OO

yx

y xlogx

y x

If f(x) = g(x),

graph y = f(x)

and y = g(x)

on the same set

of axes.

4 Use the logarithmic function log on your graphics calculator

to solve the following equations correct to 3 significantfigures. You may need to use the instructions on page 15.

628 Logarithms (Chapter 31)

For example, if f(x) = log2 x then f1(x) = 2x.

The inverse function y = log2 x is the reflection of y = 2x

in the line y = x.

a log10 x = 3 x b log10(x 2) = 2xc log10

x4

= x2 2 d log10 x = x 1

e log10 x = 5x f log10 x = 3

x 3

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Y:\HAESE\IGCSE01\IG01_31\628IGCSE01_31.CDR Tuesday, 18 November 2008 11:12:59 AM PETER

Consider two positive numbers x and y. We can write them both with base a: x = ap and y = aq, forsome p and q.

) p = loga x and q = loga y ...... (*)

Using exponent laws, we notice that: xy = apaq = ap+q

x

y=

ap

aq= apq

xn = (ap)n = anp

) loga(xy) = p+ q = loga x+ loga y ffrom *gloga

x

y

= p q = loga x loga yloga(x

n) = np = n loga x

loga(xy) = loga x + loga y

loga

x

y

= loga x loga y

loga(xn) = n loga x

Example 5 Self Tutor

If log3 5 = p and log3 8 = q, write in terms of p and q:

a log3 40 b log3 25 c log3

64125

a log3 40

= log3(5 8)= log3 5 + log3 8

= p+ q

b log3 25

= log3 52

= 2 log3 5

= 2p

c log3

64125

= log3

82

53

= log3 8

2 log3 53= 2 log3 8 3 log3 5= 2q 3p

RULES FOR LOGARITHMS [3.10]C

629Logarithms (Chapter 31)

Example 4 Self Tutor

Simplify: a log2 7 12 log2 3 + log2 5 b 3 log2 5

a log2 7 12 log2 3 + log2 5= log2 7 + log2 5 log2 3

1

2

= log2(7 5) log2p3

= log2

35p3

b 3 log2 5

= log2 23 log2 5

= log285

= log2(1:6)

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Y:\HAESE\IGCSE01\IG01_31\629IGCSE01_31.CDR Tuesday, 18 November 2008 11:13:47 AM PETER

EXERCISE 31C

1 Write as a single logarithm:

a log3 2 + log3 8 b log2 9 log2 3 c 3 log5 2 + 2 log5 3d log3 8 + log3 7 log3 4 e 1 + log3 4 f 2 + log3 5g 1 + log7 3 h 1 + 2 log4 3 3 log4 5 i 2 log3 m+ 7 log3 nj 5 log2 k 3 log2 n

2 If log2 7 = p and log2 3 = q, write in terms of p and q:

a log2 21 b log237

c log2 49 d log2 27

e log279

f log2(63) g log2

569

h log2(5:25)

3 Write y in terms of u and v if:

a log2 y = 3 log2 u b log3 y = 3 log3 u log3 vc log5 y = 2 log5 u+ 3 log5 v d log2 y = u+ v

e log2 y = u log2 v f log5 y = log5 ug log7 y = 1 + 2 log7 v h log2 y =

12 log2 v 2 log2 u

i log6 y = 2 13 log6 u j log3 y = 12 log3 u+ log3 v + 14 Without using a calculator, simplify:

alog2 16

log2 4b

logp 16

logp 4c

log5 25

log515

d logm 25logm

15

Logarithms in base 10 are called common logarithms.

y = log10 x is often written as just y = log x, and we assume the logarithm has base 10.

Your calculator has a log key which is for base 10 logarithms.

Discovery Logarithms#endboxedheading

The logarithm of any positive number can be evaluated using the log key on your calculator. You will

1 Copy and complete: Number Number as a power of 10 log of number

10

100

1000

100 000 105 log(100 000) = 5

0:1

0:001

LOGARITHMS IN BASE 10 [3.10]D

need to do this to evaluate the logarithms in this discovery.

630 Logarithms (Chapter 31)

What to do:

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Y:\HAESE\IGCSE01\IG01_31\630IGCSE01_31.CDR Friday, 31 October 2008 9:46:05 AM PETER

2 Copy and complete: Number Number as a power of 10 log of numberp10

3p10

p1000

1p10

3 Can you draw any conclusion from your table? For example, you may wish to comment on when

a logarithm is positive or negative.

Example 6 Self Tutor If the base for alogarithm is not

given then we

assume it is 10.

a 2 b 20

a b

RULES FOR BASE 10 LOGARITHMS

These rules

correspond

closely to the

exponent laws.

log(xy) = logx+ log y

log

x

y

= logx log y

log(xn) = n logx

Example 7 Self Tutor

Write as a single logarithm:

a log 2 + log 7 b log 6 log 3 c 2 + log 9 d log 49log17

a log 2 + log 7

= log(2 7)= log 14

b log 6 log 3= log

63

= log 2

c 2 + log 9

= log 102 + log 9

= log(100 9)= log 900

dlog 49

log17

=

log 72

log 71

=2 log 7

1 log 7= 2

The rules for base logarithms are clearly the same rules for general logarithms:10

631Logarithms (Chapter 31)

Use the property a = 10log a to write the following numbersas powers of 10:

log 2 0:301) 2 100:301

log 20 1:301) 20 101:301

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Y:\HAESE\IGCSE01\IG01_31\631IGCSE01_31.CDR Thursday, 30 October 2008 12:05:29 PM PETER

EXERCISE 31D.1

1

a 8 b 80 c 800 d 0:8 e 0:008

f 0:3 g 0:03 h 0:000 03 i 50 j 0:0005

2 Write as a single logarithm in the form log k:

a log 6 + log 5 b log 10 log 2 c 2 log 2 + log 3d log 5 2 log 2 e 12 log 4 log 2 f log 2 + log 3 + log 5g log 20 + log(0:2) h log 2 log 3 i 3 log 18j 4 log 2 + 3 log 5 k 6 log 2 3 log 5 l 1 + log 2

m 1 log 2 n 2 log 5 o 3 + log 2 + log 73 Explain why log 30 = log 3 + 1 and log(0:3) = log 3 14 Without using a calculator, simplify:

alog 8

log 2b

log 9

log 3c

log 4

log 8d

log 5

log15

e

log(0:5)

log 2f

log 8

log(0:25)g

log 2b

log 8h

log 4

log 2a

5 Without using a calculator, show that:

a log 8 = 3 log 2 b log 32 = 5 log 2 c log17

= log 7

d log14

= 2 log 2 e logp5 = 12 log 5 f log 3

p2 = 13 log 2

g log

1p3

= 12 log 3 h log 5 = 1 log 2 i log 500 = 3 log 2

6 74 = 2401 2400Show that log 7 34 log 2 + 14 log 3 + 12 .

LOGARITHMIC EQUATIONS

The logarithm laws can be used to help rearrange equations. They are particularly useful when dealing with

exponential equations.

Example 8 Self Tutor

Write the following as logarithmic equations in base 10:

a y = a3b2 b y =mpn

a y = a3b2

) log y = log(a3b2)

) log y = log a3 + log b2

) log y = 3 log a+ 2 log b

b y =mpn

) log y = log

m

n1

2

) log y = logm logn 12) log y = logm 12 logn

632 Logarithms (Chapter 31)

Write as powers of 10 using a = 10log a:

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Y:\HAESE\IGCSE01\IG01_31\632IGCSE01_31.CDR Thursday, 30 October 2008 11:59:02 AM PETER

Example 9 Self Tutor

Write these equations without logarithms:

a logD = 2x+ 1 b logN 1:301 2x

a logD = 2x+ 1

) D = 102x+1

or D = (100)x 10

b logN 1:301 2x) N 101:3012x

) N 101:301

102x 20

102x

Example 10 Self Tutor

Write these equations without logarithms:

a logC = log a+ 3 log b b logG = 2 log d 1

a logC = log a+ 3 log b

= log a+ log b3

= log(ab3)

) C = ab3

b logG = 2 log d 1= log d2 log 101

= log

d2

10

) G =

d2

10

EXERCISE 31D.2

1 Write the following as logarithmic equations in base 10:

a y = ab2 b y =a2

bc y = d

pp

d M = a2b5 e P =pab f Q =

pm

n

g R = abc2 h T = 5

rd

ci M =

ab3pc

2 Write these equations without logarithms:

a logQ = x+ 2 b log J = 2x 1 c logM = 2 xd logP 0:301 + x e logR x+ 1:477 f logK = 12x+ 1

3 Write these equations without logarithms:

a logM = log a+ log b b logN = log d log ec logF = 2 log x d logT = 12 log p

e logD = log g f logS = 2 log bg logA = logB 2 logC h 2 log p+ log q = log si log d+ 3 logm = logn 2 log p j logm 12 logn = 2 logP

k logN = 1 + log t l logP = 2 log x

633Logarithms (Chapter 31)

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Y:\HAESE\IGCSE01\IG01_31\633IGCSE01_31.CDR Monday, 27 October 2008 3:02:10 PM PETER

We have already seen how to solve equations such as 2x = 5 using technology. We now consider analgebraic method.

By definition, the exact solution is x = log2 5, but we need to know how to evaluate this number.

We therefore consider taking the logarithm of both sides of the original equation:

log(2x) = log 5

) x log 2 = log 5 flogarithm lawg) x =

log 5

log 2

We conclude that log2 5 =log 5

log 2.

In general: the solution to ax = b where a > 0, b > 0 is x = loga b =log b

log a:

Example 11 Self Tutor

Use logarithms to solve for x, giving answers correct to 3 significant figures:

a 2x = 30 b (1:02)x = 2:79 c 3x = 0:05

a 2x = 30

) x =log 30

log 2

) x 4:91

b (1:02)x = 2:79

) x =log(2:79)

log(1:02)

) x 51:8

c 3x = 0:05

) x =log(0:05)

log 3

) x 2:73

Example 12 Self Tutor

Show that log2 11 =log 11

log 2. Hence find log2 11.

Let log2 11 = x

) 2x = 11

) log(2x) = log 11

) x log 2 = log 11

) x =log 11

log 2

EXPONENTIAL AND LOGARITHMICEQUATIONS [3.10]

E

) log2 11 =log 11

log 2 3:46

634 Logarithms (Chapter 31)

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Y:\HAESE\IGCSE01\IG01_31\634IGCSE01_31.CDR Monday, 27 October 2008 3:02:12 PM PETER

To solve logarithmic equations, we can sometimes write each side as a power of 10.

Example 13 Self Tutor

Solve for x: log3 x = 1

log3 x = 1)

log x

log 3= 1 f loga b =

log b

log ag

) log x = 1 log 3) log x = log(31)

) log x = log13

) x = 13

EXERCISE 31E

1 Solve for x using logarithms, giving answers to 4 significant figures:

a 10x = 80 b 10x = 8000 c 10x = 0:025

d 10x = 456:3 e 10x = 0:8764 f 10x = 0:000 179 2

2 Solve for x using logarithms, giving answers to 4 significant figures:

a 2x = 3 b 2x = 10 c 2x = 400

d 2x = 0:0075 e 5x = 1000 f 6x = 0:836

g (1:1)x = 1:86 h (1:25)x = 3 i (0:87)x = 0:001

j (0:7)x = 0:21 k (1:085)x = 2 l (0:997)x = 0:5

3 The weight of bacteria in a culture t hours after it has been established

is given by W = 2:5 20:04t grams.After what time will the weight reach:

a 4 grams b 15 grams?

4 The population of bees in a hive t hours after it has

been discovered is given by P = 5000 20:09t.After what time will the population reach:

a 15 000 b 50 000?

5 Answer the Opening Problem on page 625.

6 Show that log5 13 =log 13

log 5. Hence find log5 13.

7 Find, correct to 3 significant figures:

a log2 12 b log3 100 c log7 51 d log2(0:063)

8 Solve for x:

a log2 x = 2 b log5 x = 2 c log2(x+ 2) = 2 d log5(2x) = 1

635Logarithms (Chapter 31)

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Y:\HAESE\IGCSE01\IG01_31\635IGCSE01_31.CDR Friday, 31 October 2008 9:46:39 AM PETER

Review set 31A#endboxedheading

1 a On the same set of axes, sketch the graphs of y = 2x and y = log2 x.

b What transformation would map y = 2x onto y = log2 x?

c State the domain and range of y = log2 x.

2 Copy and complete:

a loga ax = b if y = bx then x = , and vice versa.

3 Find the value of:

a log2 16 b log313

c log2

p32 d log4 8

4 Write the following in terms of logarithms:

a y = 5x b y = 7x

5 Write the following as exponential equations:

a y = log3 x b T =13 log4 n

6 Make x the subject of:

a y = log5 x b w = log(3x) c q =72x

3

7 Find the inverse function, f1(x) of:

a f(x) = 4 5x b f(x) = 2 log3 x8 Solve for x, giving your answers correct to 5 significant figures:

a 4x = 100 b 4x = 0:001 c (0:96)x = 0:013 74

9 The population of a colony of wasps t days after discovery is given by P = 400 20:03t:a How big will the population be after 10 days?

b How long will it take for the population to reach 1200 wasps?

10 Write as a single logarithm:

a log 12 log 2 b 2 log 3 + log 4 c 2 log2 3 + 3 log2 511 Write as a logarithmic equation in base 10:

a y =a3

b2b M = 3

qab

12 Write as an equation without logarithms:

a logT = x+ 3 b logN = 2 log c log d13 If log2 3 = a and log2 5 = b, find in terms of a and b:

a log2 15 b log2123

c log2 10

14 Find y in terms of u and v if:

a log2 y = 4 log2 u b log5 y = 2 log5 v c log3 y = 12 log3 u+ log3 v15 Find log3 15 correct to 4 decimal places.

16 Use a graphics calculator to solve, correct to 4 significant figures:

a 2x = 4 3x b log x = 3x

636 Logarithms (Chapter 31)

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Y:\HAESE\IGCSE01\IG01_31\636IGCSE01_31.CDR Monday, 27 October 2008 3:02:19 PM PETER

Review set 31B#endboxedheading

1 a On the same set of axes, sketch the graphs of y = 3x and y = log3 x.

b State the domain and range of each function.

2 Find the value of:

a log2p2 b log2

1p8

c logp3 27 d log9 27

3 Write the following in terms of logarithms:

a y = 4x b y = an

4 Write the following as exponential equations:

a y = log2 d b M =12 loga k

5 Make x the subject of:

a y = log3 x b T = logb(3x) c 3t = 5 2x+1

6 Find the inverse function f1(x) of:

a f(x) = 6x b f(x) = 12 log5 x

7 Solve for x, giving your answers correct to 4 significant figures:

a 3x = 3000 b (1:13)x = 2 c 2(2x) = 10

8

a What was the value of the banknote in 1970?

b What was the value of the banknote in 2005?

c

9 Write as a single logarithm:

a log2 5 + log2 3 b log3 8 log3 2 c 2 log 5 1 d 2 log2 5 110 Write as a logarithmic equation in base 10:

a D =100

n2b G2 = c3d

11 Write as an equation without logarithms:

a logM = 2x+ 1 b logG = 12 log d 112 If log3 7 = a and log3 4 = b, find in terms of a and b:

a log347

b log3 28 c log3

73

13 Find y in terms of c and d if:

a log2 y = 2 log2 c b log3 y =13 log3 c 2 log3 d

14 Find log7 200 correct to 3 decimal places.

15 Use a graphics calculator to solve, correct to 4 significant figures:

a 3x = 0:6x+ 2 b log(2x) = (x 1)(x 4)

637Logarithms (Chapter 31)

The value of a rare banknote has been modelled by V = 400 20:15t US dollars, where t is thetime in years since 1970.

When is the banknote expected to have a value of $100 000?

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Y:\HAESE\IGCSE01\IG01_31\637IGCSE01_31.CDR Friday, 31 October 2008 9:48:35 AM PETER

Challenge#endboxedheading

1 Where is the error in the following argument?12 >

14

) log(12 ) > log(14)

) log(12 ) > log(12)

2

) log(12 ) > 2 log(12 )

) 1 > 2 fdividing both sides by log(12 )g2 Solve for x:

a 4x 2x+3 + 15 = 0 Hint: Let 2x = m, say.b log x = 5 log 2 log(x+ 4).

3 a Find the solution of 2x = 3 to the full extent of your calculators display.

b The solution of this equation is not a rational number, so it is irrational.

Consequently its decimal expansion is infinitely long and neither terminates nor recurs. Copy

and complete the following argument which proves that the solution of 2x = 3 is irrationalwithout looking at the decimal expansion.

Proof:

Assume that the solution of 2x = 3 is rational.

(The opposite of what we are trying to prove.)

) there exist positive integers p and q such that x =p

q, q 6= 0

Thus 2pq

= 3

) 2p = ::::::

and this is impossible as the LHS is ...... and the RHS is ...... no matter what values p and qmay take.

Clearly, we have a contradiction and so the original assumption is incorrect.

Consequently, the solution of 2x = 3 is ......

4 Prove that:

a the solution of 3x = 4 is irrational

b the exact value of log2 5 is irrational.

638 Logarithms (Chapter 31)

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Y:\HAESE\IGCSE01\IG01_31\638IGCSE01_31.cdr Tuesday, 4 November 2008 12:05:20 PM PETER