Logrithms. Logrithm is nother word for n inde or power. THIS IS A POWER STATEMENT BASE POWER FOR EXAMPLE : We lred know tht; = NUMBER 10² = 100 This is the POWER Sttement OR 2 is the power to which the bse 10 must be rised to give 100. 2 is the logrithm which, with bse 10, gives 100. ANY POWER STATEMENT CAN BE SWOPPED AROUND TO MAKE A LOG STATEMENT log 00 2 10 1 = This is the LOGARITHMIC sttement THE TWO STATEMENTS SAY THE SAME THING BUT REARRANGED!!! Similrl if, OR 2³ = 8 (Power sttement) 3 is the power to which the bse 2 must be rised to give 8. log2 8 = 3 (Log sttement) Logrithm is nother word for power or inde. POWER FUNCTIONS re often clled EXPONENTIALS. A LOGARITHMIC FUNCTION is therefore the INVERSE (opposite) of n EXPONENTIAL function. This smbol mens These sttements cn be bbrevited s follows; IMPLIES THAT Both ws mens ech sttement implies tht the 10² = 100 log 10 100 = 2 other is true lso. 2³ = 8 log 2 8 = 3
It follows tht log 5 25 = 2 log 8 512 = 3 log 3 81 = 4 5 ³ = 125 9 ½ = 3 2 5 = 32 You will need these lter in the proofs of the Lws of Logs log = m log = n m + n = m - n = = nm n Even though we hve onl considered certin bses so fr, the bse of logrithm cn be n positive number (). Generll we cn s If log b = c c = b In the C2 emintion we concentrte on logrithms of generl bse, which could be ANY number. However our clcultors re progrmmed with the bse of 10 nd the bse of ver specil number clled e (more of e in the C3 emintion). We will lern nd prove THREE ver importnt lws of logrithms nd how these cn be used to help us solve equtions with the unknown vrible s POWER.
The Lws of Logrithms. SINCE A LOGARITHM IS A POWER THE LAWS OF LOGS SHOULD BE THE SAME AS THE LAWS OF INDICES!! AND THEY ARE! The following rules ppl to the logrithm to ANY BASE Cll it bse. You hve to KNOW them nd PROVE them!!!!! RULE 1: log + log = log RULE 2: log - log = log n Rule 3 log = nlog The following flow digrm shows the steps to ech of the three proofs of the lws of logs. Although there re three rules to prove the follow similr formt nd rel on ou being ble to swop between log sttements nd power sttements START WITH A LOG STATEMENT (OR LINK THIS TWO LOG TO THE STATEMENTS) INITIAL Let m= log STATEMENT And ou re And n = log DONE!!! QED Use the DEFINITION OF LOGS to chnge the POWER STATEMENTS BA CK into LOG sttements Use the DEFINITION OF LOGS to chnge the LOG STATEMENTS into POWER sttements m Let = n And = Use the RULES OF INDICES relevnt to the proof (dd, subtrct or multipl) Now MULTIPLY DIVIDE or RAISE TO A POWER OF n (depending on wht ou wnt to prove!)
RULE 1: Proof. Addition of logrithms (sme bse) log + log = log THE PROOF IS VERY OFTEN EXAMINED Given tht nd re positive. Let log = m nd log = n (first sttement) Therefore m = nd n = (from the definition of logrithms). It follows tht b multipling Using rules of indices (Add powers) (From the definition of logrithms) ( m )( n ) = m + n = m + n = log But from our first sttement m=log nd n=log, therefore m+n = log + log = log so log + log = log the result is proven (QED) PRACTICE WRITING IT OUT FOR YOURSELF
RULE 2: Subtrction of logs (sme bse) log - log = log Given tht nd re positive Proof. THE PROOF IS VERY OFTEN EXAMINED! Let log = m nd log = n (first sttement) Therefore m = nd n = (from the definition of logrithms). It follows tht b DIVIDING m = n B lws of indices m - n = (subtrct Powers) (from the definition of logrithms). But from the first sttement therefore m - n = log log = m nd log = n, m - n = log - log m - n = log - log = log so log - log = log nd the result is proven (QED) Tr writing it out ourself
log of power of (THE POWER RULE-COME ON DOWN!!!) n Rule 3 log = nlog Given tht is positive = m first sttement Let log Then b definition of logs m = (s before!) Now Rise both sides to the power of n (where n is nome non zero number) n m n B lws of indices = ( ) = nm = n nm now use the definition of logs to turn the power sttement bck to log sttement log n = nm b definition using the first sttement m= log n gives nm = n log = log n so log = nlog nd the result is proven (QED) Tr writing it out ourself
USING LAWS OF LOGS TO SIMPLIFY or MANIPULATE. EXERCISE Epress s single logrithm using pproprite lws of logs.: 1. Log3+log7 2. Log4+log5 3. 3log4 4. 2log3+3log2 5. 2log11+log1 6. log6-log3 7. 2log8-3log2 8. 3log5+4log6-3log10 WJEC PAST PAPER QUESTION Epress s single logrithm 1 log36 + log 256 2log 48 2 2007 JAN C2 WJEC PAST PAPER QUESTIONS (2004 OLD P2 JANUARY) Given tht is positive, simplif 1 log 6 log 3 4log 2 + WJEC PAST PAPER QUESTIONS (2003 OLD P2 JUNE) Given tht 4 log + 3log log16 = log log128 Epress in terms of. WJEC PAST PAPER QUESTIONS (2005 C2 JUNE) 3 Epress log10 2 + 2 log10 18 log10 36 s single logrithm in its simplest form 2
Solving Equtions Using Logrithms. We cn solve equtions using logs when we hve the condition tht the unknown is in the POWER. EXAMPLE 1 SOLVE 5 = 20 B tking the logs of both sides, we get log 5 = log 20 It follows tht using the POWER RULE log 5 = log 20 We cn now isolte the unknown s follows: = log 20 log 5 =1.861 (3dp) NOTE YOU CAN NOT CANCEL OUT THE WORD log!!!!!!! EXAMPLE 2 Solve the Eqution 4 3 5 = 8 correct to 3 deciml plces
EXAMPLE 3 The following emple does not hve the unknown s power but the eqution itself involves logs. A Simple cse would involve onl one log, in which cse we would use ANTILOG or in other words tke or powers (or eponents ) of both sides. This will depend on the bse being used. If we hve log (remember this mens bse 10) we will hve 10 s the bse but lter on in C3 we will hve ln (which mens log to the bse of tht specil number e) we will hve e s the bse. SOLVE Log(+3)=2 We will mke ech side s power with the bse of 10 10 = 10 log( + 3) 2 BUT BY DEFINITIONTHE LEFT HAND SIDE POWER AND LOG CANCELS OUT! 10 = 10 log( + 3) 2 ( + 3) = 100 = 97 IF THERE ARE MORE THAN ONE LOGARITHM WE CAN NOT DO THIS UNTIL WE HAVE USED LAWS OF LOGS TO MAKE ONLY ONE LOGARITHM EXAMPLE 4 Given tht log + log (3+ 4) = 2log (3 4) Where is greter thn 4, FIND THE VALUE of 3
EXERCISE SOLVING EQUATIONS WITH LOGS(nd other PPQu s) ANSWERS WJEC PAST PAPER QUESTIONS (2004 OLD P2 JUNE) Given tht 2 1 7 = 3, find the vlue of correct to three deciml plces. WJEC PAST PAPER QUESTIONS (2003 OLD P2 JANUARY) Solve the eqution 2 log( + 14) log = 2log 3 WJEC PAST PAPER QUESTIONS (2005 C2 JANUARY) Use the substitution 2 + 2 3 = u to solve the eqution 3 3 + 14=0 giving our nswers correct to three deciml plces
A CLOSER LOOK AT POWER FUNCTIONS (OR EXPONENTIAL FUNCTIONS) nd the reltionship to the log function. Let us look more crefull t the grph of = 2 Finding some vlues nd plotting some points will give us n ide of wht the grph looks like. Wht bout = 5 or = 10? We see the grph simpl gets steeper the lrger the BASE vlue.
So we notice tht lthough the grphs slightl chnge s the vlue of chnges, the ALL PASS THROUGH ONE COMMON POINT!!! This is the point (0, 1) nd it is becuse n number to the power of zero is equl to 1. Hence feture of ANY EXPONENTIAL (POWER) grph is to go through the point (0,1)
If we sketch grphs of logrithms to different bses we see there is reltionship between them We notice tht the grphs ll pss through the point (1,0) This is b definition of logrithm. The power tht n bse must be rised to in order to chieve 1 is the power of zero. So the log to the bse of nthing of 1 is zero log 0 = 1 becuse 0 = 1
Nturl Logrithms Although Nturl Logrithms do not pl prt in the C2 course I will mention them here. There is n importnt irrtionl number denoted b e. It is pproimtel equl to 2.71828 The reson for its importnce is tht if we DIFFERENTIATE (find the grdient of the tngent to n point) the eqution of the curve =e d then we get e d = number such tht if we differentite the power function we get the SAME function.. So e is the onl This vlue pls n importnt role in the modelling of popultion growth or decline, lso it cn be used to mesure the rte of dec of rdio-ctive mteril. When e is used s the bse for logrithms the re clled nturl logrithms, nd to void n confusion the re referred to s ln. Generll if, ln = e = It is importnt to relise tht ln nd e re inverse functions, this cn be shown grphicll s follows. e ln = NB. ln 1 = 0 nd e 0 = 1 ( nthing to the power of zero is equl to 1) A LOGARITHM CAN NEVER BE A NEGATIVE NUMBER
Evluting Logrithms with clcultor. Logs cn be evluted mentll if the powers nd bses re convenient, eg. Wht is the vlue of log 4 64? 4 3 = 64 log 4 64 = 3 When the re more wkwrd we use the clcultor,there re n INFINITE number of bses we could use but the Scientific clcultor will hve TWO bses on TWO different buttons. Neither button mentions the bse but the bse is implied in the following w: log refers to the logrithm to the bse of 10 ln refers to the logrithm to the bse of e (remember log is in bse 10.) LOOK FOR THESE BUTTONS ON YOUR CALCULATOR! eg. log 2 = 0.301029995... so 10 0.301029995.. = 2 log 3=0.477121254. so 10 0.477121254.. = 3 (or lmost!!!!) C2 Emintion content onl uses logs to the bse of 10 But ll proofs in C2 will be to the bse of (n bse) REMEMBER THE LAWS OF LOGS WORK FOR ANY BASE
SUMMARY OF IMPORTANT POINTS A Logrithm is POWER REMEMBER LOGS OF NEGATIVE NUMBERS DO NOT EXIST!!! This is becuse there does not eist number tht ou cn use s power tht would result in NEGATIVE number (if the BASE is positive) Also log 0 1 = becuse 0 = 1 no mtter wht the vlue of Becuse logs functions nd eponentil (power) functions re INVERSE Functions the will cncel ech other out log log VERY IMPORTANT IS HOW WE SWITHCH BETWEEN POWER STATEMENTS AND LOG STATEMENTS. = = If log b = c c = b