Logarithms
16 janv. 2001 (d) 2 log10 5 + log10 4 = log10 (52) + log10 4 = log10(25 × 4). = log10 100 = log10 (102) = 2 log10 10 = 2. (e) 3 loga 4 + loga(1/4) ? 4 loga 2 ...
a. log10 100 b. log25 5
Example 2: Write each equation in its equivalent logarithmic form. a. 26 = x b. b4 = 81 c. 2y = 128. Example 3: Evaluate each of the following. a. log10 100.
CONTRIBUTION A LETUDE DE LA QUALITE BACTERIOLOGIQUE
En ce qui concerne les coliformes totaux (CT) la concentration moyenne est de l'ordre de 1
Exercices sur le logarithme décimal
log10 a. (b) log10 µ10a3b?2 a?a2b3 ¶3 µ a?4b3. 100 4. ?b2a¶. ?2. = 3 log10. 10a3b?2 a?a2b3 ? 2 log10 2 log10 a?4 ? 2 log10 b3 + 2 log10 100 +.
What is a logarithm ?
log10 100 = 2. This is read as 'log to the base 10 of 100 is 2'. These alternative forms are shown in Figure 1. log10 100 = 2. 100 = 102 base index or power.
Exercices sur les logarithmes
d) log10. (?. 10) = 1. 2 e) log10 (100000) = 5 f) log10 (0000001) = ?5 100. ) o) 2log10. ( 1. ?. 100. ) +log10 (100).
Logarithms
log10(1000) – log10(100) = 3 – 2 = 1 = log10(10). 1000 ÷ 100 = 10. Subtract on the log scale ? divide on the natural scale. Logarithms. 100 = 1.
LES LOGARITHMES
Remarque : La suite située à gauche des flèches (100 101
Passive Intermodulation (PIM) in In-Building Distributed Antenna
7 août 2016 .01 W = 10*LOG10 (.01/.001) = 10*LOG10 (10). = 10*1.0 = 10 dBm .1 W = 10*LOG10 (.1/.001) = 10*LOG10 (100). = 10*2.0 = 20 dBm.
RMT TD n°2 Interprétation tests de croissance
24 mars 2010 soit 1 + 0.88 = 1.88 log10 cfu/g (= 76 cfu/g). - Le seuil de 100 ufc/g à durée de vie sera-t-il respecté ? oui (= 76 cfu/g < 100 cfu/g).
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; which is the result we wanted. Solutions to Problems 26Problem 1.First of all, by rule 3, we have 3log32 = log3(23) = log38. Thus the expression becomes
log38log34 + log312
log38 + log312
log34: Using rule 1, the rst expression in the [ ] brackets be- comes log 3 812= log 34:
The expression then simplies to
log34log34 = 0:
Solutions to Problems 27Problem 2.
First we use rule 3:
3log105 = log1053
and 5log102 = log1025:
Thus 3log105 + 5log102 = log1053+ log25= log105325;
where we have used rule 1 to obtain the right hand side. Thus 3log105 + 5log102log104 = log105325log104
and, using rule 2, this simplies to log1053254
= log10103= 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 2log34 + 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 5log36(2log34 + log318) = log365log34218
= log3654218
= log325354229
= log 333= 3log
33 = 3;
since log33 = 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 3log4p3 = 3log4
312=32log43; log
46 = log4(23) = log42 + log43:
Putting all of this together:
3log4(p3)12log43 + 3log42log46
32log4312log43 + 3log42(log42 + log43)
32121log
43 + (31)log42
= 2log42 = 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[PDF] logarithme base 10
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