Section 4.4 Logarithmic Properties
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1 Secion. Logarihmic Properies 5 Secion. Logarihmic Properies In he previous secion, we derived wo imporan properies of arihms, which allowed us o solve some asic eponenial and arihmic equaions. Properies of Logs Inverse Properies: Eponenial Propery: r A r A Change of Base: c ( A) A ( ) c While hese properies allow us o solve a large numer of prolems, hey are no sufficien o solve all prolems in eponenial and arihmic equaions. Properies of Logs Sum of Logs Propery: A C ( AC) Difference of Logs Propery: A A C C As an imporan noe, he arihm represens a funcion and does no follow regular algeraic disriuion rules ha you may e used o. The word does no disriue ino parenhesis, and so you mus learn hese new rules. To help in his process we offer a proof o help solidify our new rules and show how hey follow from properies you ve already seen. Le a A and c C, so y definiion of he arihm, a A and c C
2 5 Chaper Using hese epressions, AC a c a c Using eponen rules on he righ, AC Taking he of oh sides, and uilizing he inverse propery of s, a c AC a c Replacinga and c wih heir definiion esalishes he resul AC A C The proof for he difference propery is very similar. Wih hese properies, we can rewrie epressions involving muliple s as a single, or a reak an epression involving a single ino epressions involving muliple s. Eample Wrie 5 8 as a single arihm. Using he sum of s propery on he firs wo erms, This reduces our original epression o 0 Then using he difference of s propery, Eample Evaluae 5 wihou a calculaor y firs rewriing as a single arihm. On he firs erm, we can use he eponen propery of s o wrie Wih he epression reduced o a sum of wo s, 5 sum of s propery 5 ( 5) (00), we can uilize he Since 00 = 0, we can evaluae his wihou a calculaor: (00) 0 Try i Now. Wihou a calculaor evaluae y firs rewriing as a single arihm 8
3 Secion. Logarihmic Properies 55 Eample Rewrie y 7 as a sum or difference of s Firs noicing we have a quoien of wo epressions, we can uilize he difference propery of s o wrie y 7 y (7) Then seeing he produc in he firs erm, we use he sum propery y (7) ( y) (7) Finally, we could use he eponen propery on he firs erm ( y) (7) ( ) ( y) (7) Ineresingly, solving eponenial equaions was no he reason arihms were originally developed. Hisorically, up unil he adven of calculaors and compuers, he power of arihms was ha hese properies allowed muliplicaion, division, roos, and powers o e evaluaed using addiion and suracion, which is much easier o compue wihou a calculaor. Large ooks of arihm values were pulished lising he arihms of numers, such as in he ale o he righ. To find he produc of wo numers, he sum of properies were used. Suppose for eample we didn know he value of imes. Using he sum propery of s ( ) () () value (value) Using he ale, ( ) () () We can hen use he ale again in reverse, looking for as he resul of he. From ha we can deermine ( ) (6) By doing addiion and he ale of s, we were ale o deermine 6. Likewise, o compue a cue roo like 8 / ( 8) 8 (8) ( ) So ()
4 56 Chaper Alhough hese calculaions are simple and insignifican hey illusrae he same idea ha was used for hundreds of years as an efficien way o calculae he produc, quoien, roos, and powers of large and complicaed numers, eiher using ales of arihms or mechanical ools called slide rules. These properies sill have pracical applicaions for inerpreing changes in eponenial and arihmic relaionships. Eample Recall ha in chemisry, ph H. If he concenraion of hydrogen ions in a liquid is douled, wha is he affec on ph? SupposeC is he original concenraion of hydrogen ions, and P is he original ph of he liquid, so P C. If he concenraion is douled, he new concenraion is C. Then he ph of he new liquid is ph C Using he sum propery of s, ph C () ( C) () ( C) Since ph P C P ( ), he new ph is P 0.0 Afer he concenraion of hydrogen ions is douled, he ph will decrease y 0.0. Log properies in solving equaions The arihm properies ofen arise when solving prolems involving arihms Eample 5 Solve ( 50 5) ( ) In order o rewrie as an eponenial, we need a single arihmic epression on he lef side of he equaion. Using he difference propery of s, we can rewrie he lef side: 50 5 Rewriing in eponenial form reduces his o an algeraic equaion
5 Secion. Logarihmic Properies 57 Solving, Try i Now. Solve ( ) ( ) More comple eponenial equaions can ofen e solved in more han one way. In he following eample, we will solve he same prolem in wo ways one using arihm properies, and he oher using eponenial properies. Eample 6a In 008, he populaion of Kenya was approimaely 8.8 million, and was growing y.6% each year, while he populaion of Sudan was approimaely. million and growing y.% each year. If hese rends coninue, when will he populaion of Kenya mach ha of Sudan? We sar y wriing an equaion for each populaion in erms of, years afer 008. Kenya( ) 8.8( 0.6) Sudan( ).( 0.) To find when he populaions will e equal, we can se he equaions equal 8.8(.6).(.) For our firs approach, we ake he of oh sides of he equaion 8.8(.6).(.) Uilizing he sum propery of s, we can rewrie each side, (8.8).6 (.). Then uilizing he eponen propery, we can pull he variales ou of he eponen ( 8.8).6 (.). Moving all he erms involving o one side of he equaion and he res of he erms o he oher side,.6. (.) (8.8) World Bank, World Developmen Indicaors, as repored on hp:// rerieved Augus, 00
6 58 Chaper Facoring ou he on he lef,.6. (.) (8.8) Dividing o solve for (.) (8.8).6..9 years unil he populaions will e equal Eample 6 Solve he prolem aove using rewriing efore aking he Saring a he equaion 8.8(.6).(.) Divide o move he eponenial erms o one side of he equaion and he consans o he oher side Using eponen rules o group on he lef, Taking he of oh sides Uilizing he eponen propery on he lef, Dividing gives years.6.
7 Secion. Logarihmic Properies 59 While he answer does no immediaely appear idenical o ha produced using he previous mehod, noe ha y using he differen propery of s, he answer could e rewrien:. 8.8 (.) (8.8).6 (.6) (.). While oh mehods work equally well, i ofen requires less seps o uilize algera efore aking s, raher han relying on properies. Try i Now. Tank A conains 0 liers of waer, and 5% of he waer evaporaes each week. Tank B conains 0 liers of waer, and 50% of he waer evaporaes each week. In how many weeks will he anks conain he same amoun of waer? Imporan Topics of his Secion Inverse Eponenial Change of ase Sum of s propery Difference of s propery Solving equaions using rules Try i Now Answers weeks
8 60 Chaper Secion. Eercises Simplify using arihm properies o a single arihm (7) y z 6. y z Use arihm properies o epand each epression 5 7. y 8. a 9 5 z c 9. a c 5 0. a 5 c. y. y. y y y y y y y
9 Solve each equaion for he variale e e. e e ( ) ( ). 6 6 ( ) ( )
Section 4.4 Logarithmic Properties
Secion. Logarihmic Properies 59 Secion. Logarihmic Properies In he previous secion, we derived wo imporan properies of arihms, which allowed us o solve some asic eponenial and arihmic equaions. Properies
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