Expnents and Lgarithms Chapter IV - Expnents and Lgarithms A. Intrductin Starting with additin and defining the ntatins fr subtractin, multiplicatin and divisin, we discvered negative numbers and fractins. With defining new ntatins built upn lder ntatins, we were able t develp a set f rules fr perfrming the basic peratins with these newly defined numbers. We develped an understanding and facility with these numbers and peratins by practicing sample prblems with them. In the traditinal treatment f expnents and their inverses - lgarithms, we d nt practice the peratins directly, but lk results up in tables. This apprach masks the full cmprehensin in a manner similar t a way the calculatr des. Culd yu give an apprximate answer t 2 333? One f the prblems yu may bserve is that up t expnents, all f ur prblems have invlved integers r ratinal numbers. B. Definitin f an expnential A number raised t a pwer is defined as the number f times yu multiply that number times ne. Thus 2 3 = 1 x 2 x 2 x 2 = 8 Yu will ntice that we used a psitinal ntatin rather than intrduce a new peratr. In prgramming a duble asterisk is used, and in sme texts a caret is used. An example is given as fllws: 2 3 =2**3 = 2"3 Let us lk at 2 3 and 3 2 : 23=1 x 2 x 2 x 2=8 3 2 =1 x 3 x 3=9. The expnential peratin is nt cmmutative as are additin and multiplicatin. The rder f perands des make a difference in the answer. What value f A will give the same value fr the base? 2A= 2 = 1x2 The value f A=l. >-. 0.. Ụ. ;z: Q What value f B will give the same expnent? B2=2 It wuld have t be a number (21/2) that when multiplied by itself wuld give us 2. Ifwe used 3 as the expnent, we wuld get a different answer (3 1/3 ). Thus nly with the expnent, d we have a unique identity element. C. Prperties f the expnent There are several interesting peratins t study with expnents. a 2 x a 3 = a x a x a x a x a =a(3+2) When we multiply like bases raised t a pwer, we add the expnents. a 5 I a 3 = a x a x a x a x a I(a x a x a) = a x a = a(5-3)=a 2 When we divide numbers with like bases, we subtract the expnents. (a 2 i = a 2 x a 2 x a 2 = a 3x2 When we raise a base that has already be raised t a pwer, we multiply the expnents. -13-
Expnents and Lgarithms By ging t the fundamental definitin f expnents, we find that we can derive special relatinships. Let us use ur existing frmula t develp new relatinships. The secret t learning expnents is t practice deriving the frmula with numbers and with algebraic symbls. Let us summarized what we have learned: a 3 = = 1 x a x a x a = a x a x a ao= 1 a' =a a- b =lia b (abt= a bxc (abcd)e= abecde Definitin Add expnents when multiplying with like bases Subtract expnents when dividing like bases A number raise t 0 is 1 One is the identity element a negative expnent inverts Expnents f expnents are mult. t:i Z ọ... ('l '"0 -e :: D. Numerical examples The best way t understand new cncepts is t use numerical examples. Let us evaluate 16 2. 75 = 1623/4. We can write this as 16 2 +1/4+112= using the expanded ntatinal equivalent f a number. Using the multiplicative/summatin rule fr expnents, this gives us 16 2 x 16 112X 16 1/4. Nw using ur knwledge f rts we get 256 x4 x 2 =1024. Fr 4.75, we get 4 112+1/4 = 4112X41/4 = (2 2)112X (22)1/4= 2 2xl12X 22xl/4 = 21 x il2=2 x1.414.. =2.828... E. The inverses Only the expnent had a unity peratr. Thus, a' = a, s if we have a 3, then what d we have t d t the 3 t make it a I? We multiply it by 113. a l = a 3 xl/3 = (a 3 )1I3. In an example, 2 3 = 8 Then 8 1/3 = (2 3 ) 1/3 = 23xl/3 = 21 = 2. This inverse has nw allwed us t use a fractin as an expnent cmpleting the use f previusly defined elements (psitive and negative integers and fractins which can als be psitive and negative). Suppse that we wanted t find 2113. We wuld have t find a number such that when we multiplied it by itself three times wuld get the number 2. 2 1/3 = (a x a x a)1/3 = a. We will get the answer by a guessing prcedure. Since 1 x 1 x 1 = 1 < 2 < 2 x 2 x 2 = 8, we guess that the number must lie between 1 and 2. -14-
Expnents and Lgarithms 1. Finding cube rt f 2 Since 1.5 x 1.5 x1.5 = 3.375 1 < a < 1.5 Since 1.2 x 1.2 x 1.2 = 1.728 1.2 < a < 1.5 Since 1.3 x 1.3 x 1.3 = 2.197 1.2 < a < 1.3 Since 1.25 x 1.25 x 1.25 =1.953.. 1.25 < a <1.3 Since 1.27 x 1.27 x 1.27 =2.048.. 1.25 < a < 1.27 Since 1.26 x1.26 x1.26=2.0003.. 1.25 < a <1.26 Since 1.259x1.259x1.259=1.995.. 1.259 < a <1.26 After seven guesses, we have fund the answer t fur significant digits. This is a very time cnsuming prcess and gives us an incentive t lk fr a better way. What is nt bvius is that we wuld have t cntinue this prcess indefinitely because we culd nt find a repeating pattern that wuld indicate that the answer is the rati (ratinal number) f tw integers. Thus we have discvered a new number - an irratinal ne, ne that can nt be represented by the rati f tw numbers. Hwever we can apprximate the answer by a rati as we have dne with the decimal 1.259 = 1+ 25911000. If we had cntinued the guessing prcess, we wuld have fund that the answer is greater than 1.259921.. 2. Apprximatins and significant figures Since irratinal numbers are never ending we indicate the is as fllws: 2112 = 1.414... Smetimes we leave ff the..., and apprximate the number as 1.414 The number 1.14146... wuld be apprximated as fllws: 1.415 We will find that we can n lnger write an exact answer and in sme cases there may be s many digits we wuld be satisfied with just a few: 234.567 = 2.35 102 We have t experiment t see hw many digits we must utilize t get the answer crrect t 3, 4, r perhaps 5 significant figures. >. 0...... Z Cl 3. Finding lg 2(10) Let us nw rephrase ur inverse questin. the answer. 2~ 10 T what pwer d we have t raise 2, t get 10 fr Since 2 3 =8 and 2 4 =16, we knw 3 < a <4. If a =3.5, then 2 3 5. =2 3 +.5= 2 3 X 2.5= 8x 2.5where we have made use f the rules fr expnents. We have made a bld assumptin that nn-integers bey the same rules as integers. With the guessing prcess, we fmd that 2. 5 =1.414, and the we calculate 8x1.414 =11.3. Thus 3 < a <3.5. Our next guess wuld be a = 3.2 which gives us 8 x 2 2 =8x1.148..=9.18 Fr this we first find the tenth rt f2 and multiply the result twice. Since 2 1=1.0717.., 1.0717x1.0717=1.148.. Calculating the hundredth rt gets smewhat nerus. Since taking the square rt f a number is relatively easy, we culd try t get the answer by using successive square rts f 2 as described in chapter II. -15-
Expnents and Lgarithms 2112 = 1.4142136 21/4 = 1.1892071 2 1/8 = 1.0905077 2 1116 = 1.0442738 We start by dividing 10 by 2 3 =8, giving us 1.25 We can divide the 1.25 by 21/4 giving us 1.0511205 We can divide 1.0511205 by 2 1/16 instead f2 1/8 fr 2 1/8 is greater than 1.0511205 Thus 10 = 2 3 X 21/4 X 2 1116= 2 3. 3 1. (Multiplicatin is the inverse f divisin) = 8 x 1.1892071 x 1.0442738 =9.93 (nt quite 10, but clse enugh) With mre calculatins, we wuld have fund that a =3.321928.. We used the rule fr the prduct f numbers with the same base t add the expnents. We als used a binary number apprach t find the lgarithm fr which we were searching. 4. The lgarithm We nw invent a special functin called the lg functin which strips ff the expnent. lg 2(8) = lg 2(2 3 ) = 3 lg 2(2 3 ) (ntice the same numbers (2) ) Thus lg 2(10) = lg 2 (23.321928)=3.321928 Since we wrte 10=2 a, lg 2(10)= a. Rewriting we have 10 = 2 a =2 lg2(10). In anther example: 16 2314 = 2 4x(23/4) = 2 II Thus lg 2(16 23/4 ) = 11. F. Lgarithms We have defined a lgarithm as a functin which strips ff the expnent. a = b lgb(a) We can develp the fllwing 7 rules fr lgarithms I. lg b (a x c) = lg b( b lgb(a)b lgb(c~ = lg b (b IOgb(a)+lgb(c)) = lgb(a)+ lgb(c) 2. lg b (a / c) = lg b( b lgb(a) / b IOgb(c~= lg b (b lgb(a) -IOgb(c))= lgb(a) - lgb(c) 3. lg b (a") = lg b ( (b lgb (a)l) = lg b ( (b c lgb (a))) = clg b (a) 4. lg b(l) = lg b (b) = 0 5. lg b (b) = lg b (b l ) = 1 Let a = bl lgbl(a) 6. lg b2(a) = lg b2(bi IOgbl(a)) Nw let = lg bl (a) lg b2(bl) = lg b2(b l ) x lg bl (a) b2= b 1 lgbl(b2) lg b2(b2)= 1 = iogb2(b1) lg bl (b2) 7. lg b2(bl)= 1/lg bl (b2) -16-
Expnents and Lgarithms Frm first principles, we are able t develp the seven rules fr lgarithms. In the previus chapter, we saw hw t find the lgarithm by a guessing prcedure. G. Calculatins T calculate the lgarithm f a number is a three step prcess. The steps are as fllws: calculate a pwers f tw table fr the lgarithmic base f interest, divide the number by successive pwers f tw, and then add these pwers f tw t calculate the value f the lgarithm. A furth step is t check the results. Let us start by finding the lg2(10) 1. Step 1- pwers f tw table Pwer f 2 value 112 1.4142135623 114 1.1892071150 118 1.0905077326 1116 1.0442737824 (Table fr finding lgs in base 2) 1132 1.0218971486 11512 1164 1.0108892860 111024 11128 1.0054299011 112048 11256 1.0027112750 1.0013547198 1.0006771306 1.0003385080 2. Step 2 - successive divides 2 3 < 10 < 24 10/2 3 = 10/8 1.25/1.1892071150 1.0511205190/1.0442737824 1.0065564574/1.0054299011 1.0011204723/ 1.0006771306 1.0004430417 / 1.0003385080 =1.25 =1.0511205190 =1.0065564574 =1.0011204723 = 1.0004430417 =1.0001044983 rt 3 114 1116 11128 111024 112048 decimal value 3.25.0625.0078125.0009765635.00048828175 :: 3. Step 3 - add the pwers f 2 ~ lg 2(10) = 3+ 114 + 1116 + 11128 + 111024 + 112048 +... = 3 659/2048 +... = 3.321....... ;Z; 4. Step 4 - check calculatins 23.217 = 23+ 1/4+ 1116+1/128+ 1/1024+ 1/2048 = 23 X 21/4 X 2 1/16 X 2'1128X 21/1024 X 2 112048 x... =8 x 1.189207 x 1.044273 x 1.005429 x 1.000677 x 1.000338 x... =10 Let us nw try finding lg 10(2). 1. Step 1- pwers f tw table Pwer f 10 Value 112 3.16227 114 1.77827 1/8 1.33352 1116 1.15478 (Table fr finding lgs in base 10) 1132 1.07460 11512 1/64 1.03663 1/1024 11128 1.01815 112048 1/256 1.00903 114096-17- 1.00450 1.00225 1.00112 1.00056
Expnents and Lgarithms 2. Step 2 - successive divides 10 0 <2 < 10 1 2/1.77827 = 1.12468 1.12468/1.07460 =1.04661 1.04661/1.03663 =1.00962 1.009621 1.00903 =1.00059 1.000591 1.00056 =1.00003 rt 1/4 1132 1164 11256 1/4096 3. Step 3 - add the pwers f 2 lg 10(2) =1/4 + 1132 + 1164 + 11256 + 114096 =.30102 4. Step 4 - check calculatins 10.30102... = 101/4+ 1/32+ 1/64+ 1/256+ 1/4096+... = 101/4X 101/32 X 10 1164 X 10 1/256X 101/4096 x... =1.77827 x 1.07460 x 1.03663 x 1.00903 x 1.00056 decimal value.25.03125.015625.00390625.000244140 Z ~ =2 lfwe had calculated pwers ften then the check calculatin wuld be as fllws: 1030102= 103 x 1000lx 10.00002 = 1.9952 x 1.0023 x 1.00004 = 1.99986.. Nte that lg2(lo)lglo(2)= 1 =3.3219 x.30102 =.999999... (almst 1) "0 '< Finding values f numbers raised t a pwer Let us nw find 324 10. 3. 324 =1.265625 x 2 8 324 3 = 1.265625 3 X 22.4 1.265625 3 X 2.4X 22 Cnvert t binary.3 =.01001100 =.25 +.03125 +.015625 =.296875.4=.011001100 =.25+.125 +.015625 +.0078125=.3984375 1.265625 Pwer f2 114 1132 1164 value 1.0606602 1.0073886 1.0036875 2. (Table fr finding lgs in base 2) 114 1.1892071 118 1.0905077 1164 1.0108893 11128 1.0054299 1.26525.296875= 1.0606602 x 1.0073886 x 1.0036875 =1.0724371 2.3984375 = 1.1892071 x 1.0905077 x 1.0108893 x1.0054299 = 1.3180796 1.26525296875 X 2..39062522 = 1.0724371 x 1.3180796 x4 = 5.6542299-18-
Expnents and Lgarithms 324 3 = 5.6645 errr =5.664-5.654 =.010 324 multiplied 2 times 4 times 8 times X=324x324 Y=X x X Z=Y x Y 104976 1.1019961 1010 1.2143953 x 1020 10 times XxZ 1.2748236 10 25 =324 10 324 3 = 5.66x1.2748 10 25 = 7.22 10 25 H.Summary a 3 = = 1 x a x a x a = a x a x a a b a C = a b + c ab/ a C = ab-c ao= 1 a' =a a- b =lia b (ab)c= a b x c (abcd)e = abecde lg b (a x c) = lgb(a)+ lgb(c) lg b (a / c) = lgb(a) - lgb(c) lg b (a") = clg b (a) lg b(1) = 0 lg, (b) =1 lg b2(a) = lg b2(bl) x lg bl (a) lg b2{bl)= 1/lg bl (b2) Definitin Add expnents Subtract expnents A number raise t 0 is 1 One is the identity element a negative expnent inverts Expnents f expnents are mult. ~ u... Z Cl I. Prblems 1. 64 15/6= 2. Lg e(2) = 3. Lg e(3) = 4. Lg e(12) = 5. 3.14 314 = 6. Derive 7 lgarithm frmula 7. Derive 8 expnential frmula 8. Find 10 1/32 9. Find 10,1 10 1 10,00110,000110 0001 10. Find lgjo(2) using table frm prblem 8. 11. Check the result frm prblem 9. This chapter is the essence f the bk in trying t give the reader a true understanding f the meaning f expnents and lgarithms in the framewrk f the peratins f additin and multiplicatin. -19-