MATHEMATICS FOR ENGINEERING BASIC ALGEBRA


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1 MATHEMATICS FOR ENGINEERING BASIC ALGEBRA TUTORIAL  INDICES, LOGARITHMS AND FUNCTION This is the oe of series of bsic tutorils i mthemtics imed t begiers or yoe wtig to refresh themselves o fudmetls. The tutoril cotis the followig. O completio of this tutoril you should be ble to do the followig. Epli the lws of Idices. Defie logrithm. Defie tilogrithm. Epli bse umbers. Epli bsic power lws. Multiply d divide usig logrithms. Roots usig logrithms. Applictios. Logrithmic grphs. D.J.Du
2 . INTRODUCTION Before we hd electroic clcultors, ccurte clcultios ivolvig multiplictio d divisio were doe with the id of logrithms, pe d pper. To do this, the logrithms of umbers hd to be looked up i tbles d dded or subtrcted. This is esier th multiplyig d dividig. The greter the ccurcy eeded, the lrger the tbles. To do less ccurte clcultios we used slide rules d these were devices bsed o logrithms. Although we do ot eed these tody, we do use logrithms widely i mthemtics s prt of the wider uderstdig of the reltioship betwee vribles so this is importt re work. I order to uderstd logrithms, it is ecessry to uderstd idices d we should strt with this.. INDICES I lgebr, wy of writig umber or symbol such s '' tht is multiplied by itself '' times is is clled the ide. For emple 3 is the shorthd for or.. to void use of the multiplictio sig. is clled the th power of. There re four lws tht help us use this to solve problems. Lw of Multiplictio. y = + y Lw of Divisio y = y Lw of Powers. y. z = +y+z otble results = = = = times = y = the = y..... times = ( ) = Lw of Roots y = / + / + /... times / = y y = ( / ) / = WORKED EXAMPLE No. 3 ( ) Simplify the followig s = s = = 6 s = 3 D.J.Du
3 SELF ASSESSMENT EXERCISE No. Simplify the followig. C = 4. 3 (C = 7 ) d F = 4 d (F = d) 3 b.b A = 6 b (A = b ) D =. 3 (D = 3. ) 3 S = (S =. ) y S = (S =. y 3 ) y 3. DEFINITION of LOGARITHM The logrithm of umber is the power to which bse umber must be rised i order to produce it. The most commoly used bse umbers re 0 d the turl umber e which hs rouded off vlue of.783 (lso kow s Nperi Logrithms). If you hve ot come cross this umber yet, do t worry bout where it comes from but you eed to kow tht it possesses specil properties. BASE 0 This is usully show s log o clcultors but more correctly it should be writte s log 0. Sice it is the most widely used, it is lwys ssumed tht log mes with bse of 0. Suppose we wt the log of 000. We should kow tht 000 = 0 3 so the log of 000 is 3. Similrly : The log of 00 is sice 0 is 000. The log of 0 is sice 0 is 0. The log of is 0 sice 0 0 is. The log of 0. is  sice 0  is 0. d so o. This is ll well d good if we re fidig the log of multiples of 0 but wht bout more difficult umbers. I geerl if y = 0 the is the log of y d without clcultors we would hve to look them up i tbles. You c use your clcultor. SELF ASSESSMENT EXERCISE No. Use your clcultor to fid the log of the followig umbers. Just eter the umber d press the log butto. 60 (.4) 70 (.84) 6 (0.778) 0. (0.30) D.J.Du 3
4 BASE e The turl umber e (.783) will ot be eplied here but it is used for very good resos. Nturl logrithms re writte s log e or more likely s l s it ppers o most clcultors. If y = e the is the turl logrithm. You get it from your clcultor by simply eterig the umber d pressig the butto mrked l. SELF ASSESSMENT EXERCISE No.3 Use your clcultor to fid the l of the followig umbers. (swers i red) 60 (.6) 70 (4.48) 6 (.79) 0. (0.693) OTHER BASES Your clcultor my llow you to fid the logrithms to other bses by progrmmig i the bse umber but this wo t be covered here. Here re some simple emples. WORKED EXAMPLE No. Fid the log of the umber 8. Sice 3 = 8 the log (8) = 3. Fid log 3 (8). Sice 3 4 = 8 the log 3 (8) = 4 SELF ASSESSMENT EXERCISE No.4 Fid the followig (swers i red) log (6) (4) log 3 (7) (3) log (6) (4) 4. ANTILOGS A Atilog is the umber tht gives us the logrithm or put other wy, the umber resultig from risig the bse umber to the power of the logrithm. For emple if the bse is 0, Atilog() = 0 = 00 Atilog(3) = 0 3 = 000 O clcultor this is usully show s 0 d is ofte the secod fuctio of the sme key s log 0. If the bse is e the for emple til(.6) = e.6 = 60 d so o. O clcultor this is the butto mrked e d is ofte the secod fuctio of the sme key s l. D.J.Du 4
5 . POWER LAWS If A = ( ) d B = ( m ) the AB = +m If is the bse of our logrithms the = log (A) d m = log (B) d log (AB)= log A+ log B This is useful becuse if we c look up the logs of umbers we c solve multiplictio problems by ddig the logs. NEGATIVE POWERS (INDICES) If y A B = m = = the log(y) = log m = log(a) log(b) = m A B WORKED EXAMPLE No.3 Solve y = (36.)(7.7) Tkig logs we hve log y = log(36.) + log(7.7) = =.8 y = tilog(.8) = 0.8 = WORKED EXAMPLE No.4 Solve y = (36.) (7.7) Tkig logs we hve log y = log(36.)  log(7.7) = = 0.34 y = tilog(0.34) = =.060 Of course we c get the sme swers o our clcultors without this process but it is very useful to chge multiplictio ito ddig d divisio ito subtrctio. SIMPLIFYING NUMBERS WITH POWERS (INDEXES) ROOTS You kow tht: A = A hece if y = A = A the log(y) = log(a) WORKED EXAMPLE No. Fid the fifth root of 600 usig logrithms. y = log y = 600 = 600 log(600) = y = tilog(0.66) = (.778) = 3.94 = 0.6 This c be doe directly o clcultor to check the swer but the bsic trsformtio is very useful i derivtios d mipultio of formule. D.J.Du
6 6. DECIBELS The rtio of two umbers c be epressed i decibels. The defiitios is G(db) = 0 log(g) where G(db) is the rtio i decibels d G is the ctul rtio. This is commoly pplied to equipmet i which there is chge i POWER such tht G = Power Out/Power i d the G is the Gi. The reso for doig this is tht if you put two such items i series the overll gi is: G(over ll) = G G Tkig logrithms G(db) = G (db) + G (db) The gis i db re the sum of the idividul gis. 7. PRACTICAL EXAMPLES OF LOGARITHMS WORKED EXAMPLE No.6 A electroic mplifier icreses the power of the sigl by fctor of 0. Wht is the gi i decibels? G(db) = 0 log 0 = 3 db The mplifier is fed ito other mplifier with gi of. Wht is the overll gi i decibels? The gi of the secod mplifier is G = 0 log = 7 db The totl gi is = 0 db Check this wy: Overll gi = 0 = 00 I decibels G = 0 log 00 = 0 db WORKED EXAMPLE No.7 π δ A well kow formul used i the lysis of dmped vibrtios is l =  δ Where d re the mplitude of two successive vibrtios d δ is the dmpig rtio. Clculte δ whe = 3 mm d = 0. mm respectively. Clculte the mplitude reductio fctor d the dmpig rtio. 3 l l l6.79 = = = 0. π δ.79 = squre both sides  δ δ 3. =  δ 3.98δ = d δ so =  δ 3.98 =.98δ = 0.07 d δ = 0.07 = 0.74 D.J.Du 6
7 WORKED EXAMPLE No.8 Whe gs is compressed from pressure volume V d temperture T to fil volume V T d temperture T the reltioship is : T V = 8 determie C V V V C T = d T = where C is costt. Give.. = () 8 C Tke logs d log(.) = C log(8) C = log(.)/log() = 0.398/0.903 = 0.44 WORKED EXAMPLE No.9 The rtio of the tesios i pulley belt is give by R = e.µ Fid the vlue of µ whe R is. = e.µ So tke turl logs d l() =.µ l(e) d by defiitio l(e) is l() =.µ =.609 µ =.609/. = SELF ASSESSMENT EXERCISE No.. A mplifier hs gi of 3. Wht is this i decibels? (Aswer db). Give y = determie the vlue of whe y = 6 d = 0 (Aswer 0.98) 3. Give = e µ fid µ. (Aswer.48) LOGARITHMIC GRAPHS Logrithms my be used to chge to simplify fuctios by chgig them ito stright lie grph lw. Cosider the fuctio y = f() = C C is costt d power. Ecept whe =, this curve whe plotted. If we tke logrithms we fid: log(y) = φ() = log(c) + log() The grph of φ() is ow stright lie lw where log(c) is the itercept d is the grdiet. This is most useful i determiig the fuctio from eperimetl dt. D.J.Du 7
8 WORKED EXAMPLE No.0 The grph shows the results of eperimet i which vribles d y re recorded d plotted. Whe log() d log(y) re plotted the stright lie grph show is produced. Determie the fuctio f(). From the stright lie grph we hve itercept of 0.7 d grdiet of ( )/ = 3 φ() = log() Tke tilogs f() = 3 WORKED EXAMPLE No. The grph shows the results of eperimet i which vribles d y re recorded d plotted s logs. Determie the fuctio f(). From the stright lie grph we hve itercept of The grdiet is ( )/() =  0. φ() = log() Tke tilogs f() = 70. D.J.Du 8
9 SELF ASSESSMENT EXERCISE No.6 Determie the fuctio f() for ech of the grphs below, Aswers f() =. d f() = D.J.Du 9
x x x x x x x x x x x x x x x x x x
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