Appendix N: Derivation of the Logarithm Change of Base Formula
We take loga of each side of this equation which gives us loga by = loga x
MATHEMATICS 0110A CHANGE OF BASE Suppose that we have
So we get the following rule: Change of Base Formula: logb a = logc a logc b. Example 1. Express log3 10 using natural logarithms. log3 10 =.
Change of Base
Precalculus: 4.3 Rules of Loagrithms Concepts: rules of logarithms
Concepts: rules of logarithms change of base
. RulesofLogarithms
Change of Base Formula.pdf
The Change of Base Formula. Use a calculator to approximate each to the nearest thousandth. 1) log3. 3.3. 2) log2. 30. 3) log4. 5. 4) log2. 2.1. 5) log 3.55.
Change of Base Formula
Logarithms - changing the base
This leaflet gives this formula and shows how to use it. A formula for change of base. Suppose we want to calculate a logarithm to base 2. The formula states.
mc logs
Change-of-Base Formula. For any logarithmic bases a and b and
Problem #1. Use your calculator to find the following logarithms. Show your work with Change-of-Base Formula. a) b). 2 log 10. 1. 3 log 9 c). 7 log 11.
Lecture
Logarithms.pdf
16/11/2017 The log is the exponent (3); the exponent is 3 because the base used was 6. ... This Law is useful for change a logarithm in any base to a ...
Logarithms
Logarithms – University of Plymouth
16/01/2001 Use of the Rules of Logarithms. 7. Quiz on Logarithms. 8. Change of Bases. Solutions to Quizzes. Solutions to Problems ...
PlymouthUniversity MathsandStats logarithms
Secondary V Videos and Notes
Proof of the logarithm change of base rule https://youtu.be/1reblXFlM6I. Logarithm properties: review https://www.khanacademy.org/math/algebra2/.
Secondary V Videos and Notes
Change of Base
Press Í. Choose SeeGraphs from the menu. This program displays the graphs of two logarithmic functions with different bases. Y1(x)
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 log3004 = log352=2log35:
Since log
35 = 1465;we have
log3005 =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, 25x= 5: Now
5 =p25 = 2512;
so that 25x= 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, x2= 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= logaMNFrom this we are able to deduce that
log aMlogaN=mn= logaMNSolutions to Problems 25Problem 1.
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 log3004 = log352=2log35:
Since log
35 = 1465;we have
log3005 =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, 25x= 5: Now
5 =p25 = 2512;
so that 25x= 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, x2= 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= logaMNFrom this we are able to deduce that
log aMlogaN=mn= logaMNSolutions to Problems 25Problem 1.
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