Logarithms – University of Plymouth
Jan 16 2001 called the logarithm of N to the base a. ... following important rules apply to logarithms. ... Proof that loga MN = loga M + loga N.
PlymouthUniversity MathsandStats logarithms
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
6.2 Properties of Logarithms
integer and rational exponents the full proofs require Calculus. Rule2
S&Z . & .
One-Switch Utility Functions and a Measure of Risk
a proof that does not use either continuity or differentiability. PROPOSITION 2. A utility function satisfies the one-switch rule if and only if it belongs to
Elementary Functions The logarithm as an inverse function
The positive constant b is called the base (of the logarithm.) Smith (SHSU) We review the three basic logarithm rules we have developed so far.
. Logarithms (slides to )
3.3 The logarithm as an inverse function
naturally flowing out of our rules for exponents. 3.3.1 The meaning of the logarithm The positive constant b is called the base (of the logarithm.).
Lecture Notes . Logarithms
Properties of Exponents and Logarithms
Properties of Logarithms (Recall that logs are only defined for positive values of x.) For the natural logarithm For logarithms base a. 1. lnxy = lnx + lny. 1.
Exponents and Logarithms
Natural Logarithm and Natural Exponential
The function ax is called the exponential function with base a. For example we can prove the first rule in the following way: ▻ ax+y = e(x+y) ln a.
SlidesL
Near-Linear Time Algorithm with Near-Logarithmic Regret Per
Sep 28 2021 logarithmic regret per switch with sub-polynomial complexity per ... Proof. The base algorithm can restart either when we are at.
Proofs and Mathematical Reasoning
may also refer to axioms which are the starting points
Proof and Reasoning
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.
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.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.
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;- logarithm base change rule proof
- log base change rule proof