Chapter 5 indices & logarithms

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Chapter 5 indices & logarithms

  1. 1. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 46CHAPTER 5- INDICES AND LOGARITHMS5.1 INDICES AND LAW OF INDICESIn Form Three, we have learned about indices and law of indices. For the early of this chapter, we willonly revise the concept about it.Example 1:Solve the equation 0255 262=− − xxSolution:xx 262552−=)26(2552xx −=xx 4122−=01242=−+ xx0)2)(6( =−+ xx6−=x or 2=xExample 2:Find the value of y that satisfies the equation 362161.)6( 2=yySolution:232661.6 =yy23266.6 =− yyLaw of indices1.nmnmaaa +=×2.nmnmaaa −=3.mnnmaa =)( ornma4.xxxabba )(=×5. 10=a6.mnnmaa )(= or )(n mnmaa =
  2. 2. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 472)3(266 =−+ yy232 =− yy2=− y2−=yExample 3:Expressmmm22)2(36 32−+ +−in the simplest form.Solution:mmm22.22.2.36 32−+= −mmm22.82.2362−+=mmm22.82.436−+=mmm2.12.82.9 −+=m2)189( −+=m2)16(=m2)2( 4=42 += mExample 4:Given the equationxxxx 21123832 ×=× −+. Show that326 =x.Solution:xxxx 231123232 ×=× −+123123322−+= xxxx1231232 −−−+= xxxx1132 −−= xxxx3.32.2 11 −−=xx−=2332Factorize 2m
  3. 3. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 48ya x= xya =logxx2.332=x)23(32×=326 =x(proof)EXERCISE 5.11. Express 12122555 −++− nnnin the simplest terms.2. Solve the following equations.(a) 322 23=+x(b) 02781 321=− −+ xx(c) 039 531=− +− xx3. Solve the following equations.(a) 481=−x(b) xx 313255 −=4. Given811273 114=× ++ pp, find the value of p.5.2 LOGARITHMS AND LAW OF LOGARITHMS1. Logarithms can be used to simplify the operations multiplication and division of large numbers. Wecan convert from index form to logarithmic form and vice versa.2. If yax= (this is in index form), then xya =log (this is in logarithmic form) where a is known as baseand x is index.3. The logarithm to the base of 10 is known as the common logarithm. xya =log is read as logarithmof y to the base a.For example, 100102= can be converted into logarithmic form which is 2100log10 = .(i) 38log2 = is 823= .(ii) 11.0log10 −= is equal to 1)101(log10 −= which is10110 1=−(ii) 6000001.0log10 −=(iv) =∝0log10
  4. 4. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 495.2.1 Law of Logarithms1. )(logloglog mnnm aaa =+Example:)1000100(log1000log100log 101010 ×=+)100000(log 10=5=2. )(logloglognmnm aaa =−Example:)10010000(log100log1000log 101010 =−)100(log 10=2=3. mnm ana loglog =Example:31010 10log1000log =10log3 10=13 ×=3=5.2.2 Change of base of Logarithmsxba =logthen xab =Take logarithm to base c on both sides,xc ab 10loglog =axbc 10loglog =abxccloglog=Substitute into ,abbccalogloglog =1212
  5. 5. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 50Example 1:Evaluate(i) 2log8(ii) 125log25Solution:(i)8log2log2log228 =32 2log1=2log312=31=(ii)25log125log125log5525 =25355log5log=5log25log355=23=Example 2:Given that xa =2log and ya =3log . Express the following in terms of x and/or y.(i) a2log(ii) 3log2Solution:(i)2logloglog2aa aa =x1=abbccalogloglog = c is the new base that we can choose
  6. 6. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 51(ii)2log3log3log2aa=xy=EXERCISE 5.2.11. Express the following equations in index form.(a) 4log3 =p (b) 25log =p (c) x=3log22. Express the following equations in logarithm form.(a) 53=y (b) y510 = (c) rnm =5.2.3 Solving problems involving laws of logarithmsExample 1:Given 7924.03log4 = and 404.17log4 = . Without using the logarithm tables, calculate(i) 75.1log4(ii) 48log7Solution:(i) )47(log75.1log 44 =4log7log 44 −=1404.1 −=404.0=(ii)7log48log48log447 =7log)34(log424 ×=7log3log4log4424 +=7log3log4log2444 +=404.17924.02 +=404.17924.2=
  7. 7. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 529889.1=Example 2:Solve the following equations:(i) 1)13(log5log 44 =−− xx(ii)218log2log4 =− xxSolution:(i) 1)13(log5log 44 =−− xx1)135(log4 =−xx14135=−xx)13(45 −= xx4125 −= xx47 =x74=x(ii)218log2log4 =− xx218log2log 4=− xx21)82(log4=x21)816(log =x212 x=2221)2()( =x4=xExample 3:Given21136log9log212log =−+ kkk . Find the value of k.Solution:21136log9log212log =−+ kkk
  8. 8. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 5321136log9log12log 2=−+ kkk23)368112(log =×k23)27(log =k2723=k323223)27()( =k23)27(=k9=kEXERCISE 5.2.21. Given 3log =px and 5log =qx , find the values of:(a) pqxlog(b)qpxlog2. Given that xp =2log and yq =2log , express the following in terms of x or/and y.(a)qp2log(b)( )4log22pq3. Solve the equation 12log)93(log 33 =−+ tt .4. Given w3333 log2log8log4log2 =−+ , find the value of w.5.3 EQUATIONS INVOLVING INDICES AND LOGARITHMS5.3.1 Solving equations involving indicesExample 1:Solve 322 2=kSolution:322 2=k162=k4±=k
  9. 9. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 54Example 2:Given that pnm6373 == . Show thatnmmnp2+=Solution:nm73 =mnmm11)7()3( =mn73 =pn637 =pn)37(7 2×=ppn 2377 ×=Substitute into ,pmnpn 2)7(77 ×=)7(772mnppn×=mnppn277+=mnppn2+=mnpmmpn2+=mnpmpn2+=npmpmn 2+=mnnmp =+ )2(nmmnp2+= (proof)121 2
  10. 10. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 555.3.2 Solving equations involving logarithmsExample 1:Solve(i) 13log)35(log 33 −=−− tt(ii) 59 3log=x.Solution:(i) 13log)35(log 33 −=−− tt1)335(log3 −=−tt13335 −=−tt31335=−tttt 3915 =−912 =t129=t43=t(ii) 59 3log=x5log9log 3log33=x5log9loglog 333 =×x5log2log 33 =×x5loglog2 33 =x5loglog 323 =x52=x5±=x0>x5=x
  11. 11. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 56Example 2:Solve the simultaneous equations.)3(log2log 22 −+= xyyx16)64(4 =Solution:yx 234)4(4 =yx 23144 =+yx 231 =+124)3(4loglog)3(log4loglog)3(log2log2222222−=−=−+=−+=xyxyxyxySubstitute into ,525524831)124(231==−=+−=+xxxxxxSubstitute 5=x into ,8122012)5(4=−=−=yyyExample 3:Show that 5log5log5=+abab ba.Solution:abaab bbb log5loglog5+=abaabbbblog5loglog5 +÷=122 12Change the log to base b
  12. 12. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 57abababbblog5loglog5 +×=abababbblog5loglog5+=ababblog5log5 +=abbabbbloglog5log5 +=abababbabbbbblog)(log5log)log(log5=+=5= (proof)Example 4:A liquid cools from its original temperature of C°80 to a temperature CT° in x minutes. Given thatXT )88.0(80= , find(a) the temperature of the liquid after 10 minutes(b) the time taken for the liquid to cool down to C°27Solution:(a)XT )88.0(80=,10=x10)88.0(80=TC°= 28.22(b) XT )88.0(80=,27=T)8027log(88.0log80log27log88.0log)88.0log(80log27log)88.0(80log27log)88.0(8027=−=+===xxXXXAdd logb b that is actually equal to 1Factorize
  13. 13. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 5888.0log4717.04717.088.0log−=−=xx5.8=x minutesEXERCISE 5.31. Solve the equation 1374 −+= nn.2. Given 72 =x, find the value of x.3. Given that 3 3log (1 ) 2[log 1]y x− = − . Express y in terms of x.4. Find the value of x given that 27log9log 33−= x .CHAPTER REVIEW EXERCISE1. Solve the equation2 122793xxx++=2. Solve the equation ( )23 4 21x−=3. Solve the equation 4log 2 log 8 1x + =4. Given that log 3 2 = m and log 3 7 = n. Express each of the following in terms of m and / or n.(a) log 3 3.5(b) log 4 75. (a) Solve the equation ( )2 13 1x −= .(b) Mr. Ammar Fauzan invested RM 5,000 in a unit trust for his new born son’s education in thefuture. The amount of the invested money will become nRM5,000(1.09) for a period of nyears.(i) Find the amount of his investment in the fifth year, correct to two decimal points.(ii) Find the minimum number of years required so that the amount of his investment is morethan RM 20,000.6. Find the value of x and of y that satisfy both the equations below.229327xy= , 4 4 4log 2 log (2 3 ) log ( 5)x y x+ − = +7. Given that qp62 = , show thatqpq−=2log3 .8. Given that 3log xp n= and nq xlog= , show that 3=pq .9. Given 3loglog2)3(log2 555 ++=+ xx , show that 09692=+− xx .
  14. 14. Additional Mathematics Module Form 4Chapter 5- Indices And Logarithms SMK Agama Arau, PerlisPage | 5910. (a) Given that 2log3=x , find the value of x9 . Hence find the value of k9 given xk += 2 .(b) Given 5022.02log =x , prove that 5066.28log =xx .11. (a) Given that px =2log and qx =7log , express 256logxx in terms of p and q.(b) Solve the equation 2log32log2 xx =+12. Without using a scientific calculator, find the value of313log75.0log5.1log2 +−13. Solve the equation 9643 22=× +xx14. Find the values of x given that 16loglog2 xx =15. Simplify xxx2132396−++.

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