MATHEMATICS 0110A CHANGE OF BASE Suppose that we have
Change of Base Formula: logb a = logc a logc b. Example 1. Express log3 10 using natural logarithms. log3 10 = ln 10 ln 3. Example 2.
Change of Base
Logarithms - changing the base
For example logarithms to the base 2 are used in communications engineering. Your calculator can still be used but you need to apply a formula for changing the
mc logs
Lesson 5-2 - Using Properties and the Change of Base Formula
Example A. Use a calculator and the Change of Base Formula to find an approximation of log 28. log 28- log 28 log 5. ≈ 2.070. Try These A.
Precalculus: 4.3 Rules of Loagrithms Concepts: rules of logarithms
Concepts: rules of logarithms change of base
. RulesofLogarithms
6.11 Notes – Change of base and log equations
Well the reason is that we cannot evaluate a logarithm like. in our heads. Change of Base Formula: loga c = Examples: 1). 4.
day notes . notes change of base keyed
Name Objective: a. Use change-of-base formula to rewrite and
In order to evaluate logarithms with other bases you need to use the change-of-base formula. Examples: Evaluate the following logarithms. a) log4 25 b) log2 12.
Properties of Logarithms
SECTION 4.4 Evaluating Logarithms and the Change-of-Base
For example with the change-of-base theorem
. evaluating logs change of base
Logarithms – University of Plymouth
16 janv. 2001 Use of the Rules of Logarithms. 7. Quiz on Logarithms. 8. Change of Bases. Solutions to Quizzes. Solutions to Problems ...
PlymouthUniversity MathsandStats logarithms
Linear Regression Models with Logarithmic Transformations
17 mars 2011 Logarithms may have other bases for instance the decimal logarithm of base 10. (The base 10 logarithm is used in the definition of the ...
logmodels
6.2 Properties of Logarithms
Your. Calculus teacher will have more to say about this when the time comes. Example 6.2.3. Use an appropriate change of base formula to convert the following
S&Z . & .
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+ logaN2: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 log315 + 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 log212log234simplify?(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 log1010 = 1.(b)Find log366:We have 6 =p36 = 3612:
Thus log
366 = log36
361212log3636 =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
= log10103= 3log1010 = 3:(d)2log105 + log104 = log1052+ log104 = log10(254)
= log10100 = 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, forexample 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 log107 = 084510:Using the above rule,
log37 =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 log37 =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 log102 = log10105
= log1010log105
= 1069897 = 030103: Then log25 =log105log102=069897030103= 232193:
Solutions to Quizzes 15Solutions to QuizzesSolution to Quiz:Using rule 1 we have
log315 + log306 = log3(1506) = log39
But 9 = 3
2so log315 + log306 = log332= 2:End Quiz
Solutions to Quizzes 16Solution to Quiz:
Using rule 2 we have
log212log234= 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 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.Copyrightc2000rhoran@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+ logaN2: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 log315 + 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 log212log234simplify?(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 log1010 = 1.(b)Find log366:We have 6 =p36 = 3612:
Thus log
366 = log36
361212log3636 =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
= log10103= 3log1010 = 3:(d)2log105 + log104 = log1052+ log104 = log10(254)
= log10100 = 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, forexample 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 log107 = 084510:Using the above rule,
log37 =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 log37 =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 log102 = log10105
= log1010log105
= 1069897 = 030103: Then log25 =log105log102=069897030103= 232193:
Solutions to Quizzes 15Solutions to QuizzesSolution to Quiz:Using rule 1 we have
log315 + log306 = log3(1506) = log39
But 9 = 3
2so log315 + log306 = log332= 2:End Quiz
Solutions to Quizzes 16Solution to Quiz:
Using rule 2 we have
log212log234= 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- log change of base formula examples