[PDF] Logarithms - University of Plymouth

Sec 3-21 Log Logic B1 Student Versionnotebook

B log10100 C log3ã C log88 The notation is a little strange, but you can see the inverse pattern of switching the inputs and outputs The next few problems will give you an opportunity to practice thinking about this pattern and possibly make a few conjectures about other patterns that you may notice with logarithms Input

21 Log Logic - A1notebook - Ms Hansen

log10100 = 2 log101000 = 3 The notation is a little strange, but you can see the inverse pattern of switching the inputs and outputs The next few problems will give you an opportunity to practice thinking about this pattern and possibly make a few conjectures about other patterns thatyou may notice with logarithms

Work on these problems, Ms Hansen will come check off your 1

B log10100 C log3ã C log88 The notation is a little strange, but you can see the inverse pattern of switching the inputs and outputs The next few problems will give you an opportunity to practice thinking about this pattern and possibly make a few conjectures about other patterns that you may notice with logarithms

38 Solving equations involving logarithms and exponentials

Example Solve the equation e3x = 14 Solution Writing e3x = 14 in its alternative form using logarithms we obtain 3x = log e 14 = 2 639 Hence x = 2 639 3 = 0 880 To solve an equation of the form 2x = 32 it is necessary to take the logarithm of both sides of

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Logarithms - University of Plymouth

Section 1: Logarithms 3 1 Logarithms (Introduction) Let aand N be positive real numbers and let N = an:Then nis called the logarithm of Nto the base a We write this as

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5 INDICES AND LOGARITHMS

5 Indices & Logarithms 2 UNIT 5 2 SIMPLE EQUATIONS INVOLVING INDICES No Example Exercise 1 Exercise 2 1 3x = 81 3x = 34 x = 4 2x = 32 x = 4x = 64 x = 2 8x = 16 (23)x = 24 23x = 24 3x = 4 x = 3 4 4x = 32

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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; which is the result we wanted. Solutions to Problems 26Problem 1.First of all, by rule 3, we have 3log32 = log3(23) = log

38. Thus the expression becomes

log

38log34 + log312

log

38 + log312

log34: Using rule 1, the rst expression in the [ ] brackets be- comes log 3 812
= log 34:

The expression then simplies to

log

34log34 = 0:

Solutions to Problems 27Problem 2.

First we use rule 3:

3log

105 = log1053

and 5log

102 = log1025:

Thus 3log

105 + 5log102 = log1053+ log25= log105325;

where we have used rule 1 to obtain the right hand side. Thus 3log

105 + 5log102log104 = log105325log104

and, using rule 2, this simplies to log

1053254

= log

10103= 3log1010 = 3:

Solutions to Problems 28Problem 3.

Dealing rst with the expression in brackets, we have log a4 + 2loga3 = loga4 + loga32= loga432; where we have used, in succession, rules 3 and 2. Now 2log a6 = loga62 so that, nally, we have 2log a6(loga4 + 2loga3) = loga62loga432 = log a62432 = log a1 = 0:

Solutions to Problems 29Problem 4.

Dealing rst with the expression in brackets we have 2log

34 + log318 = log342+ log318 = log34218;

where we have used rule 3 rst, and then rule 1. Now, using rule 3 on the rst term, followed by rule 2, we obtain 5log

36(2log34 + log318) = log365log34218

= log

3654218

= log

325354229

= log 333
= 3log

33 = 3;

since log

33 = 1:

Solutions to Problems 30Problem 5.

The rst thing we note is that

p3 can be written as 312:We rst simplify some of the terms. They are 3log

4p3 = 3log4

312
=32log43; log

46 = log4(23) = log42 + log43:

Putting all of this together:

3log

4(p3)12log43 + 3log42log46

32log4312log43 + 3log42(log42 + log43)

32121
log

43 + (31)log42

= 2log

42 = log422= log44 = 1:

Solutions to Problems 31Problem 1.

Letx= logabandy= logbc:Then, by the denition of logarithms, a x=bandby=c:

This means that

c=by= (ax)y=axy; with the last equality following from the laws of indices. Sincec=axy; by the denition of logarithms this means that log ac=xy= logablogbc:quotesdbs_dbs47.pdfusesText_47