Core Logrithms nd eponentils Setion : Introdution to logrithms Notes nd Emples These notes ontin subsetions on Indies nd logrithms The lws of logrithms Eponentil funtions This is n emple resoure from MEI s Integrl online mthemtis resoures whih over A level Mthemtis nd Further Mthemtis for AQA, Edeel, OCR, OCR(MEI) nd WJEC. As well s notes, eerises, online tests nd intertive resoures, Integrl ontins mn resoures nd tivities to filitte nd enourge lssroom disussion nd group work. more smples nd further detils Solving eponentil equtions using logrithms An old prtil pplition of logrithms (etension work) Indies nd logrithms The importnt thing to remember bout logrithms is tht, lthough the pper to be new topi, the re simpl bout writing wht ou lred know bout indies in different w. If ou find it diffiult to work out the mening of sttement involving logrithms, it n be simpler to hnge the sttement into the equivlent sttement involving indies. log b = = b. To remember this, notie tht is both the bse of the logrithm nd the bse of the inde, nd, the logrithm, is the inde. The vlue of log b is the nswer to the question: Wht power must I rise to in order to get b? Emple (i) Find log4 (ii) Find, where log5 = Solution (i) The sttement log4 = is equivlent to 4 =. Sine 4 =, then must be. So log4 =. (ii) The sttement log5 = is equivlent to So =. 5 5 =. of 6 08/0/ MEI
C Logs Notes nd Emples You n lso look t the Flsh resoure Bsi logrithms. For prtie in understnding sttements involving logrithms, tr the intertive questions Bses of logrithms. The lws of logrithms The lws of logrithms re: log + log = log log log = log n log = nlog These n be proved using the lws of indies: It s worthwhile eerise to tr to work through these proofs First onvert into inde nottion: log log = = b = b = To prove the first lw: Similrl for the seond lw: b b + b = = = log = + b log = log + log b = log = b log = log log Using the lws of indies For the third lw: log = = n = n n log = n nlog = nlog = log n As the first two lws of indies require the indies to hve the sme bse, then the first two lws of logrithms require the logrithms to hve the sme bse. of 6 08/0/ MEI
C Logs Notes nd Emples Emple (i) Write log z in terms of log, log nd log z. (ii) Write log log b log s single logrithm. Solution (i) log = log + log log z z = log + log log z = log + log log z (ii) log log b log = log log b log log (log b log ) = + = log b You n lso look t the Flsh resoure Lws of logrithms, nd ou m find the Mthentre video Logrithms useful. For some prtie in using the lws of logrithms, tr the Logrithms puzzle. Eponentil funtions An eponentil funtion is n funtion of the form show some different eponentil funtions. =. The grphs below = 5 = =.5 of 6 08/0/ MEI
C Logs Notes nd Emples Eponentil funtions re the inverse of logrithm funtions: the funtion = is the inverse of the funtion = log. This reltionship is ver useful for solving equtions, s shown below. You n eplore grphs of this tpe using the Flsh resoure The grph of = k. Solving eponentil equtions using logrithms Logrithms re ver useful for solving equtions where the unknown vrible is n inde, suh s the eqution 0 =. Mn equtions re solved using inverse funtions, for emple ou solve the eqution + = 5, in whih ddition is pplied to the unknown vrible, b subtrting from eh side. Similrl ou solve the eqution ² = 0 b using the squre root funtion, whih is the inverse of the squre funtion. An eqution like = 0 involves n eponentil funtion of. So to solve this eqution, it follows tht ou need to use the inverse of the eponentil funtion, whih is the logrithm funtion. This is shown in the net emple. Emple Solve the following equtions. (i) = 0 (ii) = 4 (iii) 0. = Solution (i) = 0 log = log0 log = log0 log0 = =. log (ii) = 4 log = log 4 ( ) log = log 4 log 4 = log log 4 = + =. log 4 of 6 08/0/ MEI
(iii) 0. = log 0. = log ( )log 0. = log log = log 0. log = =.4 log 0. C Logs Notes nd Emples You n see similr emples using the Flsh resoure Solving ^ = b. For prtie in this tpe of problem, tr the intertive questions Logrithms: Solving powers. An old prtil pplition of logrithms Before lultors eisted, logrithms were used to mke lultions esier. For emple, suppose ou hd to divide 467 b 9675. You ould do this b long division, but it would tke long time nd the hnes of mking mistke would be quite high. So ou would ppl the seond lw of logrithms: log (467 9675) = log 467 log 9675 To do the lultion, ou would hve to find the log to bse 0 of the two numbers, subtrt the results, nd then find the inverse log of the nswer. You would hve to find the vlues of log 467 nd log 9675 from book of tbles. Unfortuntel most tbles would onl tell ou the vlues of log for vlues of between 0 nd 99. So ou would then use the ft tht log 467 = log(4.67 00000) = log4.67 + log00000 = log4.67 + 5 You would then use the tbles to find the vlue of log 4. (whih is s urte s most tbles would give ou). This would give vlue for log 467 of 6.56. You would then go through similr proess to find log 9675. log 9675 = log 96.75 + log0000 =.9855 + 4 = 5.9855 Net ou would subtrt these two logrithms (without lultor of ourse!), giving 0.707. 5 of 6 08/0/ MEI
C Logs Notes nd Emples Now ou would hve to find the number whose logrithm is 0.707. Inverse log tbles usull give vlues between nd 0. 0.707 =.707 = log 4.8 log 0 4.8 log 0 log.48 So 467 9675 =.48. = = Most pupils did not understnd the theor behind these lultions; the just followed set of instrutions to use the tbles of logrithms nd work out the lultion. Even fter lultors beme widel vilble, it ws severl ers before this tehnique ws removed from emintion sllbuses! Logrithms were lso the bsis of slide rules, whih were lso used before lultors eisted to work out lultions quikl. 6 of 6 08/0/ MEI