1.3 The log laws. Inverse functions. We have defined logx as the index needed to write x as a power of 10. For example:

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1 1.3 The log lws Inverse functions. We hve defined log s the inde needed to write s power of. For emple: since 1.8 = we hve log66.07 = 1.8 Note tht wht this sys is tht the functions f() = nd g() = log re inverse functions: mps 1.8 into nd log mps into 1.8. Putting this into digrm: : log: Tht s the ig ide: tht tking logs is the inverse of eponentition. nother wy to epress this is to write log =. If you put positive numer into your clcultor nd thn hit it with the LOG utton nd the utton, you ll get ck gin. This is lovely section. It hs ig ide (log nd ep re inverses) nd ig technique (crving log lws out of ep lws) nd when these re put together we get some cool formule. So there s lot going on nd tht cn e intimidting. Here s some good dvice: sty close to the numers. Notice how I hve done tht. Insted of using too mny symols right wy, I hve focused ttention on the reltionship etween the numers nd The log lws. Whenever new function comes into your life, one of the first things you wnt to know is how does it work? Wht cn I do with it? Wht properties does it hve? In prticulr, we re lwys interested in lgeric properties s these llow us to rewrite epressions nd solve equtions. For emple, re there ny nice lws round? Wht cn you sy out log(+y), log(y), log( r ), etc. Well, let s use the ig ide ove: log is the inverse of. We hve nice lgeric lws for the eponent lws. If we invert these we d surely get some nice lgeric lws for log. Let s try it. Emple 1. (The product lw) Tke the ddition lw for eponents: + = Wht log lw cn you etrct from tht? How should your thinking run? Well, you hve to keep hold of the ig picture. The two functions we re working with here, nd log, re inverses. Tht mens tht the inputs for one re going to e the outputs for the other. In the lw of eponents tht I wrote ove, the nd re the inputs for so they will e outputs for log. However, the log lws we re looking for will hve to involve the inputs of log nd these won t e nd, rther they ll e the outputs of, which re nd. So, to effectively use these s inputs, we hve to give them nmes nd those nmes won t e nd. Tht s the key first step, nd tht s how the following rgument hs to egin. This is n ecellent prolem to try to get the clss to tke on, ll working together with the techer coordinting from the front. Students find this tsk difficult. In mny wys, however, the prolem is essentilly nottionl. For emple if I strt them off using nd s rguments for, they ll wnt to use nd s rguments for log() s well, nd tht will surely give them grief. The key is to choose good nottion. log lws 9/13/007 1
2 The rgument. The ddition lw for eponents is + = nd t the right I hve put it in o. I like this o representtion ecuse it emphsizes the function cting s trnsformtion nd tht gets you thinking: if I do such nd such to the inputs, wht hppens to the outputs? In this cse, the o sys tht if the inputs re dded, the outputs re multiplied. Now the function log() is the inverse of, so it cn use the sme digrm ut the rrows should e reversed. Tht mens tht the ojects on the right re the inputs so we need to give them nmes. Let s set: = = B. The new version of the o is the second one t the right, nd underneth tht is the log() o with the rrows going the other wy. The first o sys tht the function mps sums into products nd the second o sys tht the function log() mps products into sums. Tht s the multipliction lw for logrithms: log(b) = log() + log(b) log( ) B B B B This twoo setup is wonderful. It encpsultes the ck nd forth etween the two functions, nd, for my money, offers just the right proof of the loglws. Emple. (The eponent lw). If we tke the product lw for logs. nd set =B, we get significnt vrint. log() = log() + log() log( ) = log(). This generlizes redily to ny positive integer, n: log( n ) = nlog(). In fct n cn e ny rel numer r. The rgument is contined in the two oes t the right. The first o displys the eponentil power lw: r = ( ) r nd the second o presents the corresponding log lw: r r log( ) r r log( r ) = rlog(). The log lws log(y) = log() + log(y) log( r ) = rlog() log(/y) = log() log(y) log(1/) = log() The lst two, the quotient lw nd the reciprocl lw, cn e derived from the first two, ut they re stted for convenience. log lws 9/13/007
3 Emple 3. Solve the eqution: 0 = 0. Solution Tke logs of oth sides: log(0 ) = log0. log(0) = log0. log0 =.37 log0 We could hve solved this eqution nd those elow in the lst section, ut with more sic pproch converting everything to se. The power lw for logrithms gives us welcome shortcut. In the previous section we did not hve the loglws ut we could still hve solved this eqution y writing everything s power of : log(0) = = log(0) = = Then the eqution ecomes: 1.30 ( ) = The eponentil power lw gives us: 1.30 = = Emple 4. Solve the eqution 3 = +1. Solution. Tke logs of oth sides: log3 = log +1 = log3 = (+1)log (log3 log)= log log log3 log Emple 5. popultion grows t the rte of 4% per dy. Over wht period will it grow y 40%? Solution. The 1dy multiplier is The dy multiplier will e (1.04). We wnt to choose so this is 1.40 (ecuse n increse of 40% is the sme multipliction y 1.40) = 1.40 Tke logs of oth sides: log(1.04 ) = log(1.40) log(1.04) = log(1.40) log1.40 = 8.58 log1.04 It will grow y 40% over period of 8.58 dys. I egin with n estimte. This will give me good check on my nswer t the end. If I put ten 4% increses together one fter the other I ll get 40% increse. Right? Wrong! Becuse of the compounding, I ll get n increse of more thn 40%. So tht tells me tht the time needed for n increse of ectly 40% is just it less thn dys nd our nswer is consistent with tht. log lws 9/13/007 3
4 Emple 6. The intensity of em of light pssing through murky wter decys with distnce trveled t constnt percentge rte. Suppose the em loses 5% every centimeter. Over wht distnce will its intensity e cut in hlf? Solution. loss of 5% is the sme s multipliction y 0.95, so this is the 1cm multiplier. Then the cm multiplier is (0.95). We wnt to choose so tht this equls (0.95) = log(0.95 ) = log0.50. log(0.95) = log0.50. = log0.50 log gin strt with n estimte. If I put ten 5% decreses together one fter the other I ll get 50% increse. Right? Wrong! Since the intensity is decresing, the loss over ech cm will decrese, nd over cm the loss will e less thn 50 cm. So tht tells me tht the distnce needed for decrese of 50% is more thn cm. The intensity will e cut in hlf over 13.5 cm. Emple 7. Solve the eqution Solution. Tke logs of oth sides: + 1 = 0. ( +1 ) log = log 0 log 0 +1 = log log = log log 0 = log I cn lmost gurntee tht hlf my clss will egin +1 = 400. Is tht vlid? Wht hve they done? They ll puse on tht, not ectly sure wht they hve done. I guess I ve squred oth sides. Relly? These prolems re esy if you follow the rules, ut follow the rules! This is good prolem in tht it forces the student to nlyze the structure of the epression. When my students first look t they see + 1 two nsties (or goodies?) squre root nd n eponentition. So there two things to e undone, ut which do we undo first? Do we squre oth sides or tke the log of oth sides? They look t me pledingly, witing to e told. I find the following conceptuliztion useful. Think of the epression + 1 s the effect of putting in sequence of oes, ech one contining the one efore. nd to identify the sequence, you imgine you + 1 hve entered in your clcultor, nd you hve to clculte y pushing the right sequence of uttons. Wht do you do? Well, you dd 1, nd then you tke the squre root nd then you eponentite with se. Ech utton corresponds to putting wht you lredy hve into new o. Now to solve the eqution you wnt to unwrp nd you do tht y opening the oes one t time. nd which o do you open first? the outside one, the lst one you closed up. So which step do you undo first? the eponentition. So the first thing to do is tke the log of oth sides. Prolems + 1 Constructing out of Step # Opertion Result 0 Begin 1 dd 1 +1 Squre root Ep (se ) + 1 log lws 9/13/007 4
5 1. Without the use of clcultor, simplify the following epressions. ll logs re se. log13 0 log13 log(0 13 ) (log0) 13 log(1/0) log4 + log5 log0 log 1/ 4 log + log(500 ). Solve the following equtions for. ll logs re se. log + 1 =. log ( +) = 1 = = 0 (log) 5 = 0 3. If log(178) =, wht is log(1.78) in terms of? If log(c) = 1.78, wht is 17.8 in terms of c? 4. The following identities re ll invlid. Mny rise from tking two trnsformtions nd pplying them one fter the other to, first in one order nd then in the other, nd setting the two results equl. This seldom works. However, it s lwys possile tht there re one or two vlues of for which the eqution just hppens to e true. In ech cse, find ll such vlues of. Your nswers should e written s simply s possile. () = () +3 = + 3 (c) 3 = 3 (d) log(3) = 3 log (e) / ( ) = / (f) = ( ) (g) log(3) = (log3)(log) (h) log(3+) = log3 + log (h) log( 3 ) = (log) 3 (i)* log( ) = log log lws 9/13/007 5
6 5.() n eponentilly growing popultion triples in size in 3 yers. How long does it tke to doule in size? () n eponentilly growing popultion doules in size every 7 dys. Wht is the dily percentge increse? How long does it tke to triple in size? To qudruple? 6. n eponentilly growing popultion increses in size y 4% every 4 hours. How long does it tke to increse in size y 8%? By %? Wht is the douling time (the time needed to doule in size)? 7. My tire hs slow lek nd loses pressure t constnt percentge rte. t noon it hs 400 kp nd t 1 pm this hs dropped to 360 kp. t wht time will it get to 0 kp? 8. ttery loses 5% of its power for every hour of use. It needs to e rechrged when its potency hs dropped to ¼ of its fully chrged vlue. For how mny hours cn it e used etween rechrgings? 9. The intensity of em of light pssing through murky wter decys with distnce trveled t constnt percentge rte. Suppose em loses 0% every meter. Over wht distnce will its intensity e cut y 90%?. In my discount wrehouse I cn sell 50 Trgiclly Hip disks per week t the regulr price of $, nd the numer of sles will increse y 8% for every % decrese in price. How mny will I sell per week if I price them t 0% off? 11. Two sunflowers grow side y side, oth incresing in height t constnt percentge rte ech dy. Sm strts t height cm nd hs growth rte 5% per dy, nd Corl strts t height 5 cm. () How long until Sm is 1 meter tll? () fter 0 dys, Sm nd Corl re the sme height. Wht is Corl s percentge growth rte? 1. For yers I've hd n eponentilly decying popultion quietly eeking out the remins of its nturl life in the hollow trunk of my old ok tree. I rememer one frosty Christms some time go when the popultion decresed y % in one week nd then y 0 g during the very net week. But just this morning I went out nd found decrese of only 8 g since my lst mesurement 7 dys go. How deep is the snow this morning? 13. n eponentilly growing popultion requires two dys to increse y 0 grms, ut needs only one dy for the net 0 grm increse. How long will it need for the third 0 grm increse? 14. decreses in vlue t the rte of % per yer wheres B decreses in vlue t the rte of 5% per yer. If strts off twice s ig s B, how long until is hlf the size of B? 15. The vlue z of n le Colville pinting increses y % every 18 months. If I plot log(z) ginst time, I will get stright line. Find the slope of this line. [Mesure time in yers nd use se for the logrithm.] 16. My ike tire hs tiny hole nd I notice tht every dy it loses % of its pressure P. If I plot log(p) ginst time, I will get stright line. Find the slope of this line. [Mesure time in dys nd use se for the logrithm.] 17. My ike tire hs tiny hole nd I notice tht the pressure P is cut in hlf every 5 dys. If I plot log(p) ginst time, I will get stright line. Find the slope of this line. [Mesure time in dys nd use se for the logrithm.] log lws 9/13/007 6
7 Multipliction through ddition The logrithm Put yourself ck in the yer You re Johnnes Kepler nd you re ttempting to clculte the orit of the plnet Venus. You hve msses of oservtionl dt to work with, collected y the dedicted efforts of your predecessor Tycho Brhe, nd you re driven y your own ingenious nd wesomely simple theoreticl constructions. But for ll tht inspirtion, the huge ulk of your work is not the hppy reordering of eutiful conceptul structures, ut hour fter hour of tedious rithmeticl mnipultions, done with s much ccurcy s the given dt will llow. Now imgine tht collegue comes long nd gives you simple scientific clcultor, nd shows you how to multiply two 6digit numers t the touch of utton. Imgine how you would feel, eing given such unimginle computtionl power. You d e lown wy, right out into the orit of one of your eloved plnets. Well tht didn t hppen, ut something just s spectculr, t lest to Kepler, did. In 1614 Scottish lnd ron nmed John Npier pulished n etensive tle which reduced the multipliction of two 7digit numers to much simpler ddition. He ws t the time 64 yers old, nd he hd spent the pst 0 yers of his life on the construction of this tle. To Kepler, this tle ws mircle nd ccording to Lplce, it proly douled the mount of scientific work he ws le to do. The ide ehind Npier s tle ws simple enough. [Interestingly enough, the ide ehind lmost ll revolutionry dvnces is simple.] Tke numer c tht s just ove 1, nd consider the sequence: 1, c, c, c 3, c 4, c 5, c 6, c 7, c 8, c 9, c, c 11, Suppose you hd two numers in tht list, sy c 3 nd c 7, nd you wnted to multiply then together. Well, if you hd possession of the list you d simply hve to know the indices tht they elonged to, 3 nd 7, nd dd them together to get, nd then look up the numer c. So possession of the tle would llow you to multiply ny two numers in the tle essentilly y dding their indices. Now if the se numer c were very close to 1, the numers c r would grow slowly nd the list would contin firly dense pcking of numers, so if you strted with ny two numers which you wnted to multiply, you could find two numers in the list which were very close, nd multiply them. Tht s essentilly how Npier s tle looked nd the reson it took him 0 yers is ecuse he hd to do ll those clcultions y hnd. n interesting wy to look t wht hppened is tht one mn did huge numer of tedious clcultions ut recorded the results in systemtic wy tht llowed ny numer of others to do their clcultions much more esily. Within few yers n importnt simplifiction (using se nd frctionl eponents) ws introduced y n English geometer Henry Briggs nd the tle ws rewritten, giving us the form of the Tle of Common Logrithms which I ws required to worked with s high school student in the 1950 s, 300 yer fter the irth of Npier.. This invention of Npier cme out of the lue. Nothing seems to hve prefigured it. But once pulished, its use spred quickly throughout the world. Within few yers it ws relized tht the ide could e incorported into mechnicl device nd hence ws orn the slide rule, which ws the hnd clcultor which we ll used s university students in the 60 s. Mine cme in lether cse with slot through which elt could pss, nd thus we crried this device round with us s we went from clss to clss just s clcultors re crried round tody. There re still lots of slide rules round in old drwers nd cupords nd oes. sk your mth techer or your prent to find you one nd see if you cn figure out wht they hve to do with logrithms. We gve tht chllenge to our students nd they found it difficult. new ending to n old joke. God sid to the cretures in the Grden, Go forth nd multiply! to which the dders replied: We cn t, we re dders. nd so tht the dders might thrive, God creted Npier. log lws 9/13/007 7
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