The document explores logarithm bases and patterns in logarithmic sequences. It finds that the value of logmn(mk) can be expressed as kn. It also determines that if loga(x)=c and logb(x)=d, then logab(x) can be calculated as cdc+d, allowing the third term in sequences to be determined.

1. Warongvat (Bank) Wanachaikiat Mathematics SL Portfolio Type I: Mathematical investigation Assignment: Logarithm Bases November 5, 2008 Logarithm Bases Introduction: A logarithm is an exponent of the power to which a base number must be raised to equal a given number. It is invented by John Napier, a Scottish mathematician who lived from 1550 to 1617. Logarithms may be defined and introduced in several ways. Logarithm of a number to a given base is the power or exponent to which the base must be raised in order to produce the number. Part 1 – Exploring logmn(mk) In this sequence, log2(8), log4(8), log16(8), log32(8) we can see that the base of this logarithm increases by the multiple of 2. However if we put the base in the form of 2n, we can see that n is increasing by 1 throughout the whole pattern. In addition, we can simplify 8 into the form of 2n which would be 23. The simplified form of this example would be: log21(23), log22(23), log23(23), log24(23), log25(23) In this form, we could deduce the pattern into a function which is: log2n(23) In this function, log2n(23): m is the constant number 2 k is the constant number 3 n is the consecutive increasing value By using the “Log Rule”, we can simplify the function into: log2n(23) = log23log2n log23log2n = 3 log2n log2 3 log2n log2 = 3n The following table was calculated by Microsoft Excel to show the solution of the example: nth termLog ofBaseEquals18232843/23883/348163/458323/568643/6781283/7882563/8985123/910810243/105081.13E+153/5010081.27E+303/100 Since we know log2n(23) = 3n, we could use it to find the solution of any nth term. If the term was 50, we could replace 50 for n and find the solution: k = 3 n = 50 log250(23) = 350 Since we know that n is the consecutive increasing value, we could find the next two terms. We also know that log2n(23) = 3n, so we could use it to find the numerical values of each terms. The next two terms of the sequence are log21(23), log22(23), log23(23), log24(23), log25(23), log26(23), log27(23) log26(23) = 36 log27(23) = 37 For the following sequences, log3(3), log9(3), log27(3), log81(3), log243(3) log5(25), log25(25), log125(25), log625(25), log3125(25) logm(mk), logm2(mk), logm3(mk), logm4(mk), logm5(mk) the next two terms are: log3(3), log9(3), log27(3), log81(3), log243(3), log729(3), log2187(3) log5(25), log25(25), log125(25), log625(25), log3125(25), log15625(25), log78125(25) logm(mk), logm2(mk), logm3(mk), logm4(mk), logm5(mk), logm6(mk), logm7(mk) The numerical values of the next two terms of each sequence are: logmn(mk) = kn log3(3), log9(3), log27(3), log81(3), log243(3), log729(3), log2187(3) log729(3) = log36(31) = 16 log2187(3) = log37(31) = 17 log5(25), log25(25), log125(25), log625(25), log3125(25), log15625(25), log78125(25) log15625(25) = log56(52) = 26 log78125(25) = log57(52) = 27 logm(mk), logm2(mk), logm3(mk), logm4(mk), logm5(mk), logm6(mk), logm7(mk) logm6(mk) = k6 logm7(mk) = k7 The following tables were calculated by Microsoft Excel to show that the answers of the function of logmn(mk) and kn are the same: nth termLog ofBaseEquals1331/12391/233271/343811/4532431/5637291/67321871/7 Functionnth termkth termEqualslog31(23)111/1log32(23)211/2log33(23)311/3log34(23)411/4log35(23)511/5log36(23)611/6log37(23)711/7 nth termLog ofBaseEquals12552/1225252/23251252/34256252/452531252/5625156252/6725781252/7 Functionnth termkth termEqualslogmn(52)122/1log52(52)222/2log53(52)322/3log54(52)422/4log55(52)522/5log56(52)622/6log57(52)722/7 nth termLog ofBaseEquals1mkm1k12mkm2k23mkm3k34mkm4k45mkm5k56mkm6k67mkm7k7 Functionnth termkth termEqualslogm1(mk)1kk1logm2(mk)2kk2logm3(mk)3kk3logm4(mk)4kk4logm5(mk)5kk5logm6(mk)6kk6logm7(mk)7kk7 The function of logmn(mk) could also be simplified by using the Change of Base rule. The simplified form is: logmn(mk) = logmklogmn logmklogmn = k logmn logm k logmn logm = kn logmn(mk) = kn The solutions of the simplified form would be the same as the solutions of logmn(mk). The following table shows the solutions to the function kn : Functionnth termkth termEqualslog21(23)133/1log22(23)233/2log23(23)333/3log24(23)433/4log25(23)533/5 This equation also applies to the following sequences: log7(7), log49(7), log343(7), log2401(7), log16807(7) log71(71), log72(71), log73(71), log74(71), log75(71) 11, 12, 13, 14, 15 log15(225), log225(225), log3375(225), log50625(225), log759375(225) log151(152), log152(152), log153(152), log154(152), log155(152) 21, 22, 23, 24, 25 Conclusion: If the logarithm appears to be in this form, logmn(mk), the numerical value would be kn . Part 2: Exploring logax, logbx, logabx In the following sequence, log4(64), log8(64), log32(64) the simplified form is: log22(26), log23(26), log25(26) log26log22, log26log23, log26log25 6 log22 log2, 6 log23 log2, 6 log25 log2 6 log22 log2, 6 log23 log2, 6 log25 log2 62, 63, 65 3, 2, 1.2 The third answer in the row can be obtained from the first two answers by multiplying the first two bases together: log4(64), log8(64) 4 × 8 32 32 is equal to the base of the third answer in the row which is log32(64). So by multiplying the first two bases of the first two answers, the third answer can be obtained. For the following rows, log7(49), log49(49), log343(49) log15125, log1125125, log1625125 log8(512), log2(512), log16(512) the third answers of each row can be calculated by multiplying bases of the first two answers together: log7(49), log49(49), log343(49) 7 × 49 = 343 log15125, log1125125, log1625125 15 × 1125 = 1625 log8(512), log2(512), log16(512) 8 × 2 = 16 This also applies to the following rows: log9(729), log27(729), log243(729) 9 ×27=243 log100(1000000), log1000(1000000), log10000(1000000) 100 ×1000=100000 If loga(x) = c and logb(x) = d, the general statement that expresses logab(x) in terms of c and d can b be found through Rules of Logarithm and Exponential Rule: loga(x) = c logb(x) = d By using the Log rule, we can simplify the equation to: ac = x bd = x then raise the equations to the power of its opposite sign: (ac)d = xd (bd)c = xc multiply both equations together: acd = xd bcd = xc acd × bcd =xc × xd (ab)cd = xc +d use the Log rule to simplify the equation: cd log(ab) = (c + d)log (x) cdc+d = log(x)log(ab) cdc+d = logab(x) The general statement that is found is cdc+d = logab(x) The two rows below can be used to test the validity of this statement: a = 36 b = 216 x = 46656 log36(46656), log216(46656), log7776(46656) log62(66), log63(66), log65(66) 62, 63, 65 3, 2, 1.2 c = 3 d = 2 3 ×23 +2= 65=1.2 a = 32768 b = 2097152 x = 64 log32768(64), log209715 (64), log68719476740(64) log85(82), log87(82), log812(82) 25, 27, 212 c = 25 d = 27 25 × 2725 + 27 = 16 Conclusion: The function of logax, logbx, logabx had been investigated through technology and using examples. As a result, the general statement of logabx is: logab(x) = cdc+d So the third answer can be calculated using this general statement. I, the undersigned, hereby declare that the following assignment is all my own work and that I worked independently on it. In this assignment, I used Microsoft Excel to draw my graphs.