Mhf4ub Unit 1 Lesson 02

2MHF4U-B

Introduction to Logarithms

Lesson 2, page 1 Advanced Functions MHF4U-B

Copyright © 2012 The Ontario Educational Communications Authority. All rights reserved. ilc.org

IntroductionScientists such as chemists, astronomers, and geologists are often confronted with either extremely large or extremely small numbers. A chemist may work with chemicals whose concentration of ions is in the realm of 0.000 000 001 or 0.000 000 1. These numbers are very difficult to compare as they stand, so you can look at them as concentrations of 10–9 or 10–7. You could say that one chemical is 100 (or 102) times as concentrated as the other. Another approach is to streamline the whole idea and say the one chemical is a factor of 2 more powerful than the other. The 2 represents the exponent on 10. This number, 2, is also called a logarithm.

A geologist may compare two earthquakes, one of which is 100 000 times as powerful as the other. She may say the first was 105 times as intense or, using the exponent (or logarithm), she may say that it was more powerful by a factor of 5.

In this unit, you will become familiar with the concept and application of logarithms.

Estimated Hours for Completing This LessonPart A: Introduction to Logarithms 1.25

Part B: Laws of Logarithms 1.25

Part C: Applications of Logarithms 1.25

Key Questions 0.75

What You Will LearnAfter completing this lesson, you will be able to

• explaintherelationshipbetweenexponentialandlogarithmicexpressions

• uselawsoflogarithmstosolvenumericexpressions

• identifykeyfeaturesoflogarithmicfunctions

• solveexponentialandlogarithmicequations



Part A: Introduction to Logarithms

Definition of Logarithm You are familiar with the concept of exponents:

23 = 8 means to multiply 2 by itself three times to get 8. (23 = 2 × 2 × 2 = 8)

You can say, “The base 2 with an exponent 3 gives 8.” Alternatively, “The exponent put on 2 to give a result of 8 is 3.” This exponent, 3, is also called a logarithm, and 23 = 8 may also be written as log28 = 3.

34 = 81 is read as “Put the exponent 4 on the base of 3 to give 81.” Alternatively, the exponent put on 3 to give a result of 81 is 4. This exponent, 4, is also called a logarithm. 34 = 81 may also be written as log381 = 4.

Generally, logax = y ay = x

Examples

Express each of the following in logarithmic form:

a) 53 = 125

b) 25 = 32

c) 70 = 1

d) 3−4

=181

e)

111

⎛⎝⎜

⎞⎠⎟

2

=1

121

f)

13

⎛⎝⎜

⎞⎠⎟

−3

= 27

g) 83= 2



Solutions

a) log5125 = 3

b) log232 = 5

c) log71 = 0

d) log3

181

= −4

e) log 1

11

1121

= 2

f) log1

3

27 = −3

g)

83 = 813

813 = 2

log8 2 =13

Examples

Express each of the following in exponential form:

a) log2128 = 7

b) log5625 = 4

c) log3

127

⎛⎝⎜

⎞⎠⎟= −3

d) log 3

4

916

⎛⎝⎜

⎞⎠⎟= 2

e) log8

12

⎛⎝⎜

⎞⎠⎟= −

13

f) log2 2 =

12

Solutions

a) 27 = 128

b) 54 = 625



c) 3−3

=127

d)

34

⎛⎝⎜

⎞⎠⎟

2

=9

16

e) 8!

13 =

12

f) 212 = 2

Examples

Evaluate each of the following:

a) log5125

b) log381

c) log2

132

⎛⎝⎜

⎞⎠⎟

d) log82

e) log 1

2

18

⎛⎝⎜

⎞⎠⎟

f) log3 27 3

g) 2log2 8

h) 7log7

149

Solutions

a) 53 = 125, so log5125 = 3

b) 34 = 81, so log381 = 4

c) 2−5

=132

, so log2

132

⎛⎝⎜

⎞⎠⎟= −5

d) 8 = 23, so 2 = 813 , so

log8 2 =

13

e)

12

⎛⎝⎜

⎞⎠⎟

3

=18

, so log 1

2

18

⎛⎝⎜

⎞⎠⎟= 3



f) 27 3 = 33( ) 3

12

= 3

3+12 = 3

72 , so

log3 27 3 =

72

g) log28 = 3, so 2log2 8 = 23 = 8

h) log7

149

= −2 , so 7

log7149 = 7−2

=149

Logarithmic EquationsThere are times in science, nature, and economics when you need to evaluate or solve a logarithmic equation for the value of a variable.

• Whatinterestratewillbeneededtoachieveamonetarygoal?

• Howmuchofanisotopeofaradioactivesubstancewaspresentinabonewhenthebodydied?

• Ifirreparabledamageisdonetoeardrumsatacertaindecibellevel, how much louder does a rock concert have to be before permanentdamageisdone?

In much the same way as you solved for unknowns with elementary equations, you can also solve for variables in logarithmic equations.

Examples

Solve for each of the following:

a) log5x = –2

b) logx49 = 2

c) log9 x =

32

d) logx2 = 3

Solutions

a) Since log5x = –2 means x = 5–2, then x =

125

.

b) Since logx49 = 2 means x2 = 49, then x = 7.



c) Since log9 x =

32

means x = 932 , then

x = 9

12

⎛

⎝⎜⎞

⎠⎟

3

= 33= 27 .

d) Since logx2 = 3 means x3 = 2, then x = 23= 2

13 .

Nature and physics do not always present numbers that are kind. That is, many of the values you work with are not integer values or powers of each other. There are times when you need to find the logarithm of a number that is not an integral power of the base.

Example

Evaluate log2100.

Solution

You are trying to solve log2100 = x or 2x = 100.

You know that 26 = 64 and 27 = 128, so 6 < log2100 < 7.

Use a bisection form of trial and error to find the value.

Halfwaybetween6and7is6.5

6 + 72

⎛⎝⎜

⎞⎠⎟.

Try 6.5: 26.5 = 90.51

This is too small.

Halfwaybetween6.5and7is6.75

6.5+ 72

⎛⎝⎜

⎞⎠⎟.

Try 6.75: 26.75 = 107.63

This is too large.

Halfwaybetween6.5and6.75is6.625

6.5+ 6.752

⎛⎝⎜

⎞⎠⎟.

Try 6.625: 26.625 = 98.70

This is too small.


6.625+ 6.752

⎛⎝⎜

⎞⎠⎟.

Try 6.6875: 26.6875 = 103.07

This is too large.




6.625+ 6.68752

⎛⎝⎜

⎞⎠⎟.

Try 6.65625: 26.65625 = 100.86

This is too large.


6.625+ 6.656252

⎛⎝⎜

⎞⎠⎟.

Try 6.640625: 26.640625 = 99.78

As you can see, the values being used are zeroing in on the actual value of log 2100 = 6.644.

Support Questions(do not send in for evaluation)

Be sure to try the Support Questions on your own before looking at the suggested answers provided.

11. Evaluate each of the following:

a) log749

b) log11121

c) log2

116

d) log3

127

e) log164

f) log2515

g) log6416

h) 2log2 16

i) log2 8 2

j) log21

11. a) log749 = 2 (since 49 = 72)

b) log11121 = 2 (since 121 = 112)

c) log2

116

= –4

116

=124 = 2−4⎛

⎝⎜⎞⎠⎟

since

116

=124 = 2−4⎛

⎝⎜⎞⎠⎟

d) log3

127

= –3

127

=133 = 3−3⎛

⎝⎜⎞⎠⎟

since

127

=133 = 3−3⎛

⎝⎜⎞⎠⎟

e) log164 = 1

2 4 = 16 =16

12

⎛

⎝⎜⎞

⎠⎟since

4 = 16 =16

12

⎛

⎝⎜⎞

⎠⎟

f) log25

15

= –1

2since

15=

125

=1

2512

= 25!

12

"

#$$

%

&''

g) log6416 = 23

(since 16 = 64x, 24 = (26)x,

24 = 26x, 4 = 6x, x = 46

= 23

)

h) 2log2 16 = 16 (since log216 = 4 because 16 = 24)

i) log2 8 2 = 72

8 2 = 23212 = 2

3+12 = 2

72

!"#

$%&since 8 2 = 232

12 = 2

3+12 = 2

72

!"#

$%&

j) log21 = 0 (since 20 = 1)



12. Solve each of the following equations:

a) log5x = 3

b) logx49 = 2

c) log9x = 32

d) log2(x – 1) = 4

13. Explain why there is no solution to the equation x = log20.

14. Use systematic trial and error to evaluate log3100.

Part B: Laws of Logarithms

Law of Logarithms for Multiplication (LLM) Remember from your laws for exponents that ax × ay = ax+y.

For example: 23 × 25 = 23+5 = 28.

You could read this in words as “the exponent you get from multiplying two numbers with the same base is the sum of the exponents of the powers.”

Expressed as a logarithm, this would be loga(p × q) = logap + logaq.

For example:

log2(8 × 32) = log2256 = 8 (since 28 = 256)

log28 + log232 = log223 + log22

5 = 3 + 5 = 8

It follows that log2(8 × 32) = log28 + log232.

14. To solve log3100 = x or 3x = 100:

You know that 34 = 81 and 35 = 243, so 4 < log3100 < 5

Try 4.5: 34.5 = 140.296, which is too big

(4 + 4.5)/2 = 4.25


(4 + 4.25)/2 = 4.125

Try 4.125: 34.125 = 92.923, which is too small

(4.125 + 4.25)/2 = 4.1875


(4.1875 + 4.25)/2 = 4.21875


(4.1875 + 4.21875)/2 = 4.203125

Try 4.203125: 34.203125 = 101.251 which is too big

(4.1875 + 4.203125)/2 = 4.1953125

You eventually come to x = 4.1918

13. x = log20 means 2x = 0. There is no exponent that can be put on the base 2 to give a value of 0 for the power. Therefore no value of x exists and so the equation cannot be solved.

12. a) log5x = 3; x = 53 = 125

b) logx49 = 2; x2 = 49, x = 7

c) log9x = 32

; x = 9

32 = 9

12

⎛

⎝⎜⎞

⎠⎟

3

= 33= 27

d) log2(x – 1) = 4; x – 1 = 24, x = 17



Examples


a) log62 + log618

b) log42 + log48

c) log102 + log1025 + log1020

Solutions

a) log62 + log618 = log6(2 × 18) = log636 = 2

b) log42 + log48 = log4(2 × 8) = log416 = 2

c) log102 + log1025 + log1020 = log10(2 × 25 × 20) = log101000 = 3

Examples


a) log3 9 3

b) log 2

3

49×

278

⎛

⎝⎜

⎞

⎠⎟

Solutions

a) log3 9 3 = log3 9+ log3 3 = 2+

12= 2

12

= 52

b) log 2

3

49×

278

⎛

⎝⎜

⎞

⎠⎟ = log 2

3

49

⎛⎝⎜

⎞⎠⎟+ log 2

3

278

⎛

⎝⎜

⎞

⎠⎟ = 2+ log 2

3

827

⎛

⎝⎜

⎞

⎠⎟

−1

= 2−32=

12

= 2+ log 23

23

3

–1

= 2+ log 23

23

– 32= 2 –

32=

12



Law of Logarithms for Division (LLD)Remember from the laws for exponents that ax ÷ ay = ax–y.

For example: 25 ÷ 22 = 25–2 = 23.

You could read this in words as “The exponent you get from dividing two numbers with the same base is the difference of the exponents of the powers.”

Expressed as a logarithm, this would be loga(p ÷ q) = logap – logaq.

For example:

log2(32 ÷ 4) = log28 = 3 (since 23 = 8)

log232 – log24 = log225 – log22

2 = 5 – 2 = 3

It follows that log2(32 ÷ 4) = log232 – log24.

Examples


a) log248 – log212

b) log3162 – log32

c) log58 – log5200

Solutions

a) log248 – log212 = log2(48 ÷ 12) = log24 = 2

b) log3 162− log3 2 = log3

1622

⎛⎝⎜

⎞⎠⎟= log3 81 = 4

c) log5 8 − log5 200 = log5

8200

⎛⎝⎜

⎞⎠⎟= log5

125

⎛⎝⎜

⎞⎠⎟= −2



Law of Logarithms for Powers (LLP)You also recall from the laws for exponents that (ax)y = axy.

For example: (73)5 = (73)(73)(73)(73)(73) = 73+3+3+3+3 = 715 = 73×5.

Writteninwords,thiswouldbe“Theexponentyougetfromraising a power to an exponent is the product of the exponents.”

In logarithmic form, this is written as logapn = nlogap.

For example:

log7493 = log7(72)3 = log77

2×3 = log776 = 6

log7493 = 3log749 = 3 × 2 = 6

This implies that log7493 = 3log749.

Examples

Express each of the following as the product of a number and a log:

a) log5103

b) log456

c) logx yz

Solutions

a) log5103 = 3log510

b) log456 = 6log45

c) logx yz = zlogx y

Examples

Evaluate the following:

a) log2 32

b) log5 253



Solutions

a) log2 32 = log2 32

12 =

12

log2 32 =12

5( ) =52

b) log5 253 = log5 25

13 =

13

log5 25 =13

2( ) =23

There are times when you may be required to use a combination of all three laws of logarithms.

Examples


a) log3 81 27

b) log5

253

125

⎛

⎝⎜

⎞

⎠⎟

Solutions

a) log3 81 27 = log3 81+ log3 27 = log3 81+ log3 27

12 = 4 +

12

log3 27 = 4 +32= 5

12

log3 81 27 = log3 81+ log3 27 = log3 81+ log3 27

12 = 4 +

12

log3 27 = 4 +32= 5

12

b)

log5

253

125

= log5 253 − log5 125 =

13

log5 25− log5 53

=13

2( )−3 =23

– 3 =23

–93= –

73= – 2

13

The Common Logarithm In some fields, people work with other bases. Computer engineers and computer scientists work in binary (base 2), which is made up of only the numbers 0 and 1. To simplify their calculations, they may sometimes work in octal (base 8), but normally find it more convenient to work in hexadecimal (base 16). (If you would like to research these bases and their uses, do an Internet search on octal or hexadecimal arithmetic and see what you can find.)

Mainstream society works almost exclusively in base 10. For this reason, you will often come up against problems where you will



be working in base 10 and therefore using log10. Since log10 is used so often, the practice is to drop the 10 from the log.

If you see, for example, log100, that is the same as log10100.

Whennobaseisindicated,base10isunderstood:

loga = log10a



a) log102.5 + log1040

b) log714 + log73.5

c) log248 – log212

d) log3162 – log32

e) log58 – log5200

f) log36 + log312 – log38


a) log2 32

b) log5 253

c) log2 4 163− log4 0.253

d) log1000 103

17. Use systematic trial and error to evaluate log7.

17. To solve log7 = x or 10x = 7:

You know that 100 = 1 and 101 = 10, so 0 < log7 < 1


(0.5 + 1)/2 = 0.75


(0.75 + 1)/2 = 0.875


(0.75 + 0.875)/2 = 0.8125


(0.8125 + 0.875)/2 = 0.84375

Try 0.84375: 100.84375 = 6.978

You eventually come to x = 0.8451.

16. a) log2 32 = log2 25( )

12 =

52

b) log5 253 = log5 52( )

13 =

23

c)

log2 4 163 − log4 0.253 = log2 22( ) 24( )

13

− log4

14

13= 2+

43−−13

=

113

log2 4 163 − log4 0.253 = log2 22( ) 24( )

13

− log4

14

13= 2+

43−−13

=

113

d) log1000 103

= log 103 ×1013

⎛

⎝⎜⎞

⎠⎟=

103

15. a) log102.5 + log1040 = log10(2.5 × 40) = log10100 = 2

b) log714 + log73.5 = log7(14 × 3.5) = log749 = 2

c) log2 48− log2 12 = log2

4812

= log2 4 = 2

d) log3 162− log3 2 = log3

1622

= log3 81 = 4

e) log5 8− log5 200 = log5

8200

= log5

125

= −2

f) log3 6+ log3 12− log3 8 = log3

6×128

= log3 9 = 2



Part C: Applications of Logarithms As you saw in Lesson 1, economic, geopolitical, scientific, and physical results can be modelled using exponential functions.

Governments need to plan for the future. Infrastructure such as hospitals, highways, sewage systems, schools, water treatment plants, electrical generating stations, seniors’ residences, and many other things must be planned for and built. To plan effectively, governments must be able to anticipate future populations.

Examples

The population of Ontario on July 1, 2006, was 12.69 million, and it was growing at a rate of 1.1% per year.

a) WhatwillthepopulationbeonJuly1,2016?

b) Whenwillthepopulationbecome15million?

Solutions

The population growth model is the same as the compound interest formula: A = P(1 + i)n

a) Since P = 12.69, i = 1.1% and n = 10 years

A = 12.69(1 + 0.011)10

A = 12.69(1.011)10

A = 12.69(1.115607836)

A = 14.16

The population will be 14.16 million people.

b) Since A = 15, P = 12.69 and i = 1.1%

15 = 12.69(1.011)n

(1.011)n

=15

12.69

(1.011)n = 1.182



At this point, two different approaches can be taken to solve for n:

i) Trial and error:

For n = 10: (1.011)10 = 1.1156

Try n = 20: (1.011)20 = 1.2446, which is too big

Try n = 15: (1.011)15 = 1.1783, which is too small

Try n = 17: (1.011)17 = 1.2044, which is too big

You would now try n = 16 and gradually narrow down the value for n until you get something you are happy with.

ii) Common logarithms:

15 = 12.69(1.011)n

By applying LLM: log15 = log12.69 + log(1.011)n

log15 – log12.69 = log(1.011)n

By applying LLP: log15 – log12.69 = nlog(1.011)

0.07263 = n(0.00475)

n = 15.29

The decimal part of 15.29 represents a fraction of a year. In months this fraction is 0.29 year × 12 months/year = 3.48 or 3 months.

Therefore the population will reach 15 million in about 15 years, 3 months, or sometime in October 2021.

The Richter scale, which measures the magnitude of earthquakes, is based on a comparison of intensities to I0, which is an earthquake of intensity 0. Since the numbers can get extremely large, a base of 10 is used. This leads to the use of logarithms base 10, or common logarithms.

The equation used is M = log

II0

⎛

⎝⎜⎞

⎠⎟.

It can be rearranged to I = I0 × 10M.



Example

Howdoestheintensityofamagnitude3.3earthquakecomparetotheintensityofamagnitude4.2earthquake?

Solution

The magnitude of the first earthquake will be M1, and that of the second one M2.

I1 = I0 ×10M1 and I2 = I0 ×10M2

I2

I1

=I0 ×10M2

I0 ×10M1=10M2 −M1 =104.2−3.3

=100.9= 7.94

The second earthquake is 7.94 times as powerful as the first.

Decibels are a measure of loudness: it is also compared to a base measure that is the minimum intensity that can be heard by a human ear. This is I0 = 10–12W/m2 (watts per square metre).

The equation used is L =10 log

II0

⎛

⎝⎜⎞

⎠⎟.

Example

A typical rock concert measures a dB level of 115, a whisper measures30dB.Howmanytimeslouderisarockconcertthanawhisper?

Solution

The intensity of the rock concert will be I1 , and the whisper I2 .

115 =10 log

I1

I0

⎛

⎝⎜⎞

⎠⎟, so

log

I1

I0

⎛

⎝⎜⎞

⎠⎟=11.5 and I1 = I0 × 1011.5

30 =10 log

I2

I0

⎛

⎝⎜⎞

⎠⎟, so

log

I2

I0

= 3 and I2 = I0 × 103

I1

I2

=I0 ×1011.5

I0 ×103=108.5

= 316 227 766

The concert is about 316 000 000 times louder than a whisper.



pHmeasurestheacidityofsubstancessuchasliquidsbyusinga logarithmic scale based on the relative acidity in moles/litre. Pure water has a concentration of 10–7 mol/L of hydrogen ions (H+).Squarebracketsareusedtoindicateconcentration.[H+] is read “the concentration of hydrogen ions.”

pH isgivenbypH=–log([H+])

Since water has a hydrogen ion concentration of 10–7 mol/L, its pHis–log10–7 or 7.

Example

LemonadehasanH+ concentration of 0.0050 mol/L.

WhatisitspH?

Solution

pH=–log0.0050

= –(–2.301)

= 2.301

ThepHofthelemonadeisabout2.3.

Example

ColashaveapHofabout2.5.WhatistheH+concentration?

Solution

2.5=–log([H+])

–2.5=log([H+])

[H+] = 10–2.5

[H+] = 0.00316

= 3.16 × 10–3

TheH+ concentration in cola is about 3.2 × 10–3 mol/L.




18. The earth’s population is approximately 6 billion people. Itisincreasingatabout1.5%peryear.Howlongwillit takefortheearth’spopulationtobecome9billion?

19. An investor wants to double his money. For how long must he investat4.5%perannumfortheoriginalamounttodouble?

20. On Sept 30, the Auckland Islands suffered two earthquakes. The first at 4:23 in the afternoon measured 7.4 on the Richter scale, whileanaftershockat8:47intheeveningmeasured6.6.Howmuchmorepowerfulwasthefirstearthquake?

21. A gas lawnmower registers at 90 dB, while a typical rock concert registers120dB.Howmuchlouderisarockconcertthanalawnmower?

22. Aflaskcontains500mLofanacidwithapHof4.If 500mLofwaterisadded,whatwillbethepHofthenewsolution?

22. ApHof4indicatesaconcentrationof10–4 or 0.0001 mol/L. Since the volume is doubled, the concentration is halved to 0.00005.

pH=–log(0.00005)=4.301

21. L1 =10 log

I1

I0

⎛

⎝⎜⎞

⎠⎟; L2 =10 log

I2

I0

⎛

⎝⎜⎞

⎠⎟; L2 − L1 =10 log

I2

I0

⎛

⎝⎜⎞

⎠⎟−10 log

I1

I0

⎛

⎝⎜⎞

⎠⎟

L1 =10 log

I1

I0

⎛

⎝⎜⎞

⎠⎟; L2 =10 log

I2

I0

⎛

⎝⎜⎞

⎠⎟; L2 − L1 =10 log

I2

I0

⎛

⎝⎜⎞

⎠⎟−10 log

I1

I0

⎛

⎝⎜⎞

⎠⎟

L2 − L1 =10 log

I2

I0

⎛

⎝⎜⎞

⎠⎟− log

I1

I0

⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟ =10 log

I2

I1

⎛

⎝⎜⎞

⎠⎟

120 − 90 =10 log

I2

I1

⎛

⎝⎜⎞

⎠⎟; 3 = log

I2

I1

⎛

⎝⎜⎞

⎠⎟;

I2

I1

=103=1000

The concert is 1000 times as loud as a lawnmower.

20.

I1

I2

=I0 ×10M1

I0 ×10M2=10M1 −M2 =107.4−6.6

=100.8= 6.31

The first earthquake was 6.31 times as powerful as the second.

19. A = P(1 + i)n; 2P = P(1 + 0.045)n; 2 = (1.045)n; log2 = nlog1.045; n = 15.7

It will take about 16 years.

18. A = P(1 + i)n; 9 = 6(1 + 0.015)n; 1.5 = (1.015)n; log1.5 = nlog1.015; n = 27.2

It will take about 27 years.



Key Questions

Now work on your Key Questions in the online submission tool. You may continue to work at this task over several sessions,

but be sure to save your work each time. When you have answered all the unit’s Key Questions, submit your work to the ILC.

(20 marks)

8. Evaluate each of the following: (6 marks)

a) log636

b) log6

136

⎛⎝⎜

⎞⎠⎟

c) 7log7 12

d) log168

e) log3 81 27

f) log1

3

1

9. Solve each of the following: (3 marks)

a) log7x = 2

b) logx64 = 3

c) log3(2x – 1) = 3

10. Evaluate each of the following: (3 marks)

a) log82 + log832

b) log5150 – log56

c) log520 + log510 – 3log52

11. AGICpays6%perannum.Howlongwouldittake$3000togrowto$6000?(3 marks)

12. Anearthquakeregisters6.1ontheRichterscale.Whatistheratingonanearthquakethatistwiceaspowerful? (3 marks)

13. Use systematic trial and error to evaluate log50. (2 marks)

http://www.ilc.org/myilc/online_submission/index.php?c_code=MHF4UB&u_no=1



Now go on to Lesson 3. Send your answers to the Key Questions to ILC when you have completed Unit 1 (Lessons 1 to 5).

Mhf4ub Unit 1 Lesson 02

Documents

Transcript of Mhf4ub Unit 1 Lesson 02