# Ib Math Portfolio Logarithms

IB Math SL

Type I Portfolio

Lisa Phommaseng

Logarithm Base

Consider the given sequences:

Log28, log48, log88, log 168, log328,…

Log381, log981, log 2781, log8181,…

Log525, log2525, log12525, log62525,…

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For the first sequence Log28, log48, log88, log 168, log328,… you are to find the next two terms of each of the sequences you would have to determine the pattern. As we can see, the value of the bases, 2,4,8,16,32 are increasing by squares of 2 (ex: 22= 4, 23=8, 24=16 etc.,). The two terms following log328 will be log648 and log1288. To find the expression for the nth term of each sequence, you would have to write the expression in the form of . First you would find log28 to begin to look for a pattern. A formula that could be useful to solve this would be logbx = . The variable b would be the base, x would be the number of the logarithm. Saying that log28 = x, we are able to determine that x is 3 because 23 equals 8. The same would follow for the next terms : log48= x, 8 is equivalent to 23 and the base of 4 is equivalent to 22 (Log28 =). So, to that term would be . You would continue on with the next and you would notice that the pattern is 3 over whatever the number that follows. Now we see that the nth term of this sequence is an=. The next sequence Log381, log981, log 2781, log8181,…you would determine the following two terms as the one above. A we can see the terms are squaring as well. The bases are increasing as so : 31=3, 32=9, 33= 27, 34=81, and so forth. We can determine that the following terms are going to be log24381 and log72981. You would also find the expression for the nth term the same way. Using the formula logbx = we are able to see that log381 = x and x = 34 or 4. Following that log981 is equal to for log981 = . The nth term of this sequence is an=. So, the following two terms of the last two sequences would be log312525 and log1562525 for the sequence Log525, log2525, log12525, log62525,…...

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