Logarithms – University of Plymouth

212085 Levelling-Up Basic MathematicsLogarithmsRobin HoranThe aim of this document is to provide a short, self assessment programme for students who wish to acquire a basic competence in the use of logarithms.Copyrightc

2000rhoran@plymouth.ac.ukLast Revision Date: January 16, 2001 Version 1.00

Table of Contents1.Logarithms2.Rules of Logarithms3.Logarithm of a Product4.Logarithm of a Quotient5.Logarithm of a Power6.Use of the Rules of Logarithms7.Quiz on Logarithms8.Change of BasesSolutions to QuizzesSolutions to Problems

Section 1: Logarithms 31. Logarithms (Introduction)LetaandNbe positive real numbers and letN=an:Thennis

called thelogarithm ofNto the basea. We write this asn= logaN:Examples 1(a)Since 16 = 24;then 4 = log216:(b)Since 81 = 34;then 4 = log381:(c)Since 3 =p9 = 912;then 1=2 = log93:(d)Since 31= 1=3;then1 = log3(1=3):

Section 1: Logarithms 4ExerciseUse the denition of logarithm given on the previous page to deter-

mine the value ofxin each of the following.1.x= log3272.x= log51253.x= log2(1=4)4.2 = logx(16)5.3 = log2x

Section 2: Rules of Logarithms 52. Rules of LogarithmsLeta;M;Nbe positive real numbers andkbe any number. Then the

following important rules apply to logarithms.1:logaMN= logaM+ logaN

2:logaMN= logaMlogaN

3:logamk=klogaM

4:logaa= 1

5:loga1 = 0

Section 3: Logarithm of a Product 63. Logarithm of a Product1. Proof thatlogaMN= logaM+ logaN:Examples 2(a)log64 + log69 = log6(49) = log636:

Ifx= log636;then 6x= 36 = 62:

Thus log

64 + log69 = 2:(b)log520 + log414= log52014:

Now 2014= 5 so log520 + log414= log55 = 1:Quiz.To which of the following numbers does the expression log

315 + log306 simplify?(a)4(b)3(c)2(d)1

Section 4: Logarithm of a Quotient 74. Logarithm of a Quotient1. Proof thatlogaMN= logaMlogaN:Examples 3(a)log240log25 = log2405= log28:

Ifx= log28 then 2x= 8 = 23;sox= 3:(b)If log35 = 1:465 then we can nd log306:

Since 3=5 = 06;then log306 = log335= log33log35:

Now log

33 = 1;so that log306 = 11465 =0465Quiz.To which of the following numbers does

the expression log

212log234simplify?(a)0(b)1(c)2(d)4

Section 5: Logarithm of a Power 85. Logarithm of a Power1. Proof thatlogamk=klogaMExamples 4(a)Find log10(1=10000):We have 10000 = 104;so 1=10000 =

1=104= 104:

Thus log

10(1=10000) = log10104=4log1010 =4;where

we have used rule 4 to write log

1010 = 1.(b)Find log366:We have 6 =p36 = 3612:

Thus log

366 = log36

3612

12log3636 =12:Quiz.If log35 = 1465;which of the following numbers is log3004?(a)-2.930(b)-1.465(c)-3.465(d)2.930

Section 6: Use of the Rules of Logarithms 96. Use of the Rules of LogarithmsIn this section we look at some applications of the rules of logarithms.Examples 5(a)log41 = 0:(b)log1010 = 1:(c)log10125 + log108 = log10(1258) = log101000

= log

10103= 3log1010 = 3:(d)2log105 + log104 = log1052+ log104 = log10(254)

= log

10100 = log10102= 2log1010 = 2:(e)3loga4+loga(1=4)4loga2 = loga43+loga(1=4)loga24

= log a4314loga24= loga42loga24 = log a16loga16 = 0:

Section 6: Use of the Rules of Logarithms 10ExerciseUse the rules of logarithms to simplify each of the following.1.3log32log34 + log3122.3log105 + 5log102log1043.2loga6(loga4 + 2loga3)4.5log36(2log34 + log318)5.3log4(p3)12log43 + 3log42log46

Section 7: Quiz on Logarithms 117. Quiz on LogarithmsIn each of the following, ndx:Begin Quiz1.logx1024 = 2(a)23(b)24(c)22(d)252.x= (logap27logap8logap125)=(loga6loga20)(a)1(b)3(c)3/2(d)-2/33.logc(10 +x)logcx= logc5)(a)2.5(b)4.5(c)5.5(d)7.5End Quiz

Section 8: Change of Bases 128. Change of BasesThere is one other rule for logarithms which is extremely useful in

practice. This relates logarithms in one base to logarithms in a dier- ent base. Most calculators will have, as standard, a facility for nding logarithms to the base 10 and also for logarithms to basee(natural logarithms). What happens if a logarithm to a dierent base, for

example 2, is required? The following is the rule that is needed.logac= logablogbc1. Proof of the above rule

Section 8: Change of Bases 13The most frequently used form of the rule is obtained by rearranging the rule on the previous page. We have log ac= logablogbcso logbc=logaclogab:Examples 6(a)Using a calculator we nd that log103 = 047712 and log

107 = 084510:Using the above rule,

log

37 =log107log103=084510047712= 177124:(b)We can do the same calculation using instead logs to basee.

Using a calculator, log

e3 = 109861 and loge7 = 194591: Thus log

37 =ln7ln3=194591109861= 177125:

The calculations have all been done to ve decimal places, which explains the slight dierence in answers. Section 8: Change of Bases 14(c)Given only that log105 = 069897 we can still nd log25;as follows. First we have 2 = 10=5 so log

102 = log10105

= log

1010log105

= 1069897 = 030103: Then log

25 =log105log102=069897030103= 232193:

Solutions to Quizzes 15Solutions to QuizzesSolution to Quiz:

Using rule 1 we have

log

315 + log306 = log3(1506) = log39

But 9 = 3

2so log

315 + log306 = log332= 2:End Quiz

Solutions to Quizzes 16Solution to Quiz:

Using rule 2 we have

log

212log234= log21234

Now we have 1234= 1243=1243= 16:

Thus log

212log234= log216 = log224:

Ifx= log224;then 2x= 24;sox= 4:End Quiz

Solutions to Quizzes 17Solution to Quiz:

Note that

004 = 4=100 = 1=25 = 1=52= 52:

Thus log

3004 = log352=2log35:

Since log

35 = 1465;we have

log

3005 =21465 =2:930:End Quiz

Solutions to Problems 18Solutions to ProblemsProblem 1. Since x= log327 then, by the denition of a logarithm, we have 3 x= 27:

But 27 = 3

3;so we have

3 x= 27 = 33; giving x= 3:

Solutions to Problems 19Problem 2.

Sincex= log255 then, by the denition of a log-

arithm, 25
x= 5: Now

5 =p25 = 2512;

so that 25
x= 5 = 2512;

From this we see thatx= 1=2:

Solutions to Problems 20Problem 3.

Sincex= log2(1=4);then, by the denition of a

logarithm, 2 x= 1=4 = 1=(22) = 22:

Thusx=2:

Solutions to Problems 21Problem 4.

Since 2 = log

x(16) then, by the denition of log- arithm, x

2= 16 = 42:

Thus x= 4:

Solutions to Problems 22Problem 5.

Since 3 = log

2x, by the denition of logarithm,

we must have 2 3=x:

Thusx= 8:

Solutions to Problems 23Problem 1.

Letm= logaMandn= logaN;so, by denition,M=amand

N=an:Then

MN=aman=am+n;

where we have used the appropriate rule for exponents. From this, using the denition of a logarithm, we have m+n= loga(MN): Butm+n= logaM+logaN;and the above equation may be written log aM+ logaN= loga(MN); which is what we wanted to prove.

Solutions to Problems 24Problem 1.

As before, letm= logaMandn= logaN:ThenM=amand

N=an:Now we have

MN=aman=amn;

where we have used the appropriate rule for indices. By the denition of a logarithm, we have mn= logaMN

From this we are able to deduce that

log aMlogaN=mn= logaMN

Solutions to Problems 25Problem 1.

Letm= logaM;soM=am:Then

M k= (am)k=amk=akm; where we have used the appropriate rule for indices. From this we have, by the denition of a logarithm, km= logaMk:

Butm= logaM;so the last equation can be written

klogaM=km= logaMk; Levelling-Up Basic MathematicsLogarithmsRobin HoranThe aim of this document is to provide a short, self assessment programme for students who wish to acquire a basic competence in the use of logarithms.Copyrightc

2000rhoran@plymouth.ac.ukLast Revision Date: January 16, 2001 Version 1.00

Table of Contents1.Logarithms2.Rules of Logarithms3.Logarithm of a Product4.Logarithm of a Quotient5.Logarithm of a Power6.Use of the Rules of Logarithms7.Quiz on Logarithms8.Change of BasesSolutions to QuizzesSolutions to Problems

Section 1: Logarithms 31. Logarithms (Introduction)LetaandNbe positive real numbers and letN=an:Thennis

called thelogarithm ofNto the basea. We write this asn= logaN:Examples 1(a)Since 16 = 24;then 4 = log216:(b)Since 81 = 34;then 4 = log381:(c)Since 3 =p9 = 912;then 1=2 = log93:(d)Since 31= 1=3;then1 = log3(1=3):

Section 1: Logarithms 4ExerciseUse the denition of logarithm given on the previous page to deter-

mine the value ofxin each of the following.1.x= log3272.x= log51253.x= log2(1=4)4.2 = logx(16)5.3 = log2x

Section 2: Rules of Logarithms 52. Rules of LogarithmsLeta;M;Nbe positive real numbers andkbe any number. Then the

following important rules apply to logarithms.1:logaMN= logaM+ logaN

2:logaMN= logaMlogaN

3:logamk=klogaM

4:logaa= 1

5:loga1 = 0

Section 3: Logarithm of a Product 63. Logarithm of a Product1. Proof thatlogaMN= logaM+ logaN:Examples 2(a)log64 + log69 = log6(49) = log636:

Ifx= log636;then 6x= 36 = 62:

Thus log

64 + log69 = 2:(b)log520 + log414= log52014:

Now 2014= 5 so log520 + log414= log55 = 1:Quiz.To which of the following numbers does the expression log

315 + log306 simplify?(a)4(b)3(c)2(d)1

Section 4: Logarithm of a Quotient 74. Logarithm of a Quotient1. Proof thatlogaMN= logaMlogaN:Examples 3(a)log240log25 = log2405= log28:

Ifx= log28 then 2x= 8 = 23;sox= 3:(b)If log35 = 1:465 then we can nd log306:

Since 3=5 = 06;then log306 = log335= log33log35:

Now log

33 = 1;so that log306 = 11465 =0465Quiz.To which of the following numbers does

the expression log

212log234simplify?(a)0(b)1(c)2(d)4

Section 5: Logarithm of a Power 85. Logarithm of a Power1. Proof thatlogamk=klogaMExamples 4(a)Find log10(1=10000):We have 10000 = 104;so 1=10000 =

1=104= 104:

Thus log

10(1=10000) = log10104=4log1010 =4;where

we have used rule 4 to write log

1010 = 1.(b)Find log366:We have 6 =p36 = 3612:

Thus log

366 = log36

3612

12log3636 =12:Quiz.If log35 = 1465;which of the following numbers is log3004?(a)-2.930(b)-1.465(c)-3.465(d)2.930

Section 6: Use of the Rules of Logarithms 96. Use of the Rules of LogarithmsIn this section we look at some applications of the rules of logarithms.Examples 5(a)log41 = 0:(b)log1010 = 1:(c)log10125 + log108 = log10(1258) = log101000

= log

10103= 3log1010 = 3:(d)2log105 + log104 = log1052+ log104 = log10(254)

= log

10100 = log10102= 2log1010 = 2:(e)3loga4+loga(1=4)4loga2 = loga43+loga(1=4)loga24

= log a4314loga24= loga42loga24 = log a16loga16 = 0:

Section 6: Use of the Rules of Logarithms 10ExerciseUse the rules of logarithms to simplify each of the following.1.3log32log34 + log3122.3log105 + 5log102log1043.2loga6(loga4 + 2loga3)4.5log36(2log34 + log318)5.3log4(p3)12log43 + 3log42log46

Section 7: Quiz on Logarithms 117. Quiz on LogarithmsIn each of the following, ndx:Begin Quiz1.logx1024 = 2(a)23(b)24(c)22(d)252.x= (logap27logap8logap125)=(loga6loga20)(a)1(b)3(c)3/2(d)-2/33.logc(10 +x)logcx= logc5)(a)2.5(b)4.5(c)5.5(d)7.5End Quiz

Section 8: Change of Bases 128. Change of BasesThere is one other rule for logarithms which is extremely useful in

practice. This relates logarithms in one base to logarithms in a dier- ent base. Most calculators will have, as standard, a facility for nding logarithms to the base 10 and also for logarithms to basee(natural logarithms). What happens if a logarithm to a dierent base, for

example 2, is required? The following is the rule that is needed.logac= logablogbc1. Proof of the above rule

Section 8: Change of Bases 13The most frequently used form of the rule is obtained by rearranging the rule on the previous page. We have log ac= logablogbcso logbc=logaclogab:Examples 6(a)Using a calculator we nd that log103 = 047712 and log

107 = 084510:Using the above rule,

log

37 =log107log103=084510047712= 177124:(b)We can do the same calculation using instead logs to basee.

Using a calculator, log

e3 = 109861 and loge7 = 194591: Thus log

37 =ln7ln3=194591109861= 177125:

The calculations have all been done to ve decimal places, which explains the slight dierence in answers. Section 8: Change of Bases 14(c)Given only that log105 = 069897 we can still nd log25;as follows. First we have 2 = 10=5 so log

102 = log10105

= log

1010log105

= 1069897 = 030103: Then log

25 =log105log102=069897030103= 232193:

Solutions to Quizzes 15Solutions to QuizzesSolution to Quiz:

Using rule 1 we have

log

315 + log306 = log3(1506) = log39

But 9 = 3

2so log

315 + log306 = log332= 2:End Quiz

Solutions to Quizzes 16Solution to Quiz:

Using rule 2 we have

log

212log234= log21234

Now we have 1234= 1243=1243= 16:

Thus log

212log234= log216 = log224:

Ifx= log224;then 2x= 24;sox= 4:End Quiz

Solutions to Quizzes 17Solution to Quiz:

Note that

004 = 4=100 = 1=25 = 1=52= 52:

Thus log

3004 = log352=2log35:

Since log

35 = 1465;we have

log

3005 =21465 =2:930:End Quiz

Solutions to Problems 18Solutions to ProblemsProblem 1. Since x= log327 then, by the denition of a logarithm, we have 3 x= 27:

But 27 = 3

3;so we have

3 x= 27 = 33; giving x= 3:

Solutions to Problems 19Problem 2.

Sincex= log255 then, by the denition of a log-

arithm, 25
x= 5: Now

5 =p25 = 2512;

so that 25
x= 5 = 2512;

From this we see thatx= 1=2:

Solutions to Problems 20Problem 3.

Sincex= log2(1=4);then, by the denition of a

logarithm, 2 x= 1=4 = 1=(22) = 22:

Thusx=2:

Solutions to Problems 21Problem 4.

Since 2 = log

x(16) then, by the denition of log- arithm, x

2= 16 = 42:

Thus x= 4:

Solutions to Problems 22Problem 5.

Since 3 = log

2x, by the denition of logarithm,

we must have 2 3=x:

Thusx= 8:

Solutions to Problems 23Problem 1.

Letm= logaMandn= logaN;so, by denition,M=amand

N=an:Then

MN=aman=am+n;

where we have used the appropriate rule for exponents. From this, using the denition of a logarithm, we have m+n= loga(MN): Butm+n= logaM+logaN;and the above equation may be written log aM+ logaN= loga(MN); which is what we wanted to prove.

Solutions to Problems 24Problem 1.

As before, letm= logaMandn= logaN:ThenM=amand

N=an:Now we have

MN=aman=amn;

where we have used the appropriate rule for indices. By the denition of a logarithm, we have mn= logaMN

From this we are able to deduce that

log aMlogaN=mn= logaMN

Solutions to Problems 25Problem 1.

Letm= logaM;soM=am:Then

M k= (am)k=amk=akm; where we have used the appropriate rule for indices. From this we have, by the denition of a logarithm, km= logaMk:

Butm= logaM;so the last equation can be written

klogaM=km= logaMk;