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IB Pre-Calculus Logarithm Bases General Information: Logarithms A logarithm is an exponent and can be described as the exponent needed to produce a certain number.

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Introduction

International Baccalaureate IB Pre-Calculus Portfolio Logarithm Bases January 6, 2009 Student Number: 1208769 C. Leon King High School Andre Elliott January 6, 2009 Dr. Stone IB Pre-Calculus Logarithm Bases General Information: Logarithms A logarithm is an exponent and can be described as the exponent needed to produce a certain number. For Example: 23=8, from this you would say that 3 is the logarithm of 8 with base of 2 (log28). 2 is written as a subscript, and 3 is the exponent to which 2 must be raised to produce 8. The formula or definition that is used in logarithms is: logbx=e, so be=x. So base(b) with exponent(e) produces x. So from the example, 2 is the base(b), 8 is x (the number produced), and the exponent(e) is 3. So, 23=8. Logarithms in Sequences: Introduction Since logarithms can be solved (log28=3) to form numbers, this means logarithms are just another way to represent a number; and since numbers can be in sequences, so can logarithms. Given this, consider the following sequences: 1) log2 8, log4 8, log8 8, log16 8, log32 8, ... 2) log3 81, log9 81, log27 81, log81 81, ... 3) log5 25, log25 25, log125 25, log625 25, ... 4) logm mk, logm2 mk, logm3 mk, logm4 mk Sequence 4 is the general statement that is used to reflect the previous sequences. So, consider the first sequence; it starts out with a base(m) ...read more.

Middle

The expression used to find the nth term (tn) of the first sequence is tn= log2n 8 = ln 8/ ln 2n. The expression used to find the nth term (tn) of the second sequence is tn= log3n 81= ln 81/ ln 3n.The expression used to find the nth term (tn) of the third sequence is tn= log5n 25 = ln 25/ln 5n. The expression used to find the nth term (tn) of the fourth sequence is tn= logmn mk = ln mk/ln mn. The answers can be easily justified by using the highly technological technology in a Texas Instruments graphing calculator. Considering log2 8=3, when plugging in ln 8/ ln2, the calculator provides the answer of 3, which is equivalent to the correct answer mentioned before. Logarithms in Sequences: Calculating Logarithms As mentioned earlier, logarithms can be solved and produce number answers, such as log28=3. Using the expressions from above this logarithm can also be solved like this: ln 8/ ln 2=3. Given this, consider the following logarithmic sequences: 1) log4 64, log8 64, log32 64 2) log7 49, log49 49, log343 49 3) log1/5 125, log1/125 125, log1/625 125 4) log8 512, log2 512, log16 512 Each of these logarithms can be calculated by using the formula logab=ln a/ ln b. In row 1, the calculated answers are as follows: ln 64/ ln 4, ln 64/ ln 8, ln 64/ ln 32. ...read more.

Conclusion

Consecutively you should get 3, 2, and 1.2. Now convert 1.2 into the fraction of 6/5. From this you can notice that 6 is made when multiplying 3 and 2, and 5 is made when adding 3 and 2. Now try another sequence. From sequence 3, plug in ln 125/ ln (1/5), ln 125/ ln (1/125), and ln 125/ ln (1/625) all separately. You should get -3, -1, and -.75 respectively. Now lets try the formula of cd/ c+d. when plugged in you should get, (-3 * -1)/ (-3 + -1), which is reduced to 3/ -4, which is equal to -.75. This proves the statement to be valid. With formulas comes a limitation. However it seems as though this one lacks a limit in terms of a, b and x. It seems this formula of cd/c+d is only useful when trying to find the calculation of a logarithm using its roots. By splitting up a logarithm into two smaller ones, you can find its solution easier by using the formula. This can be helpful when used in certain situations; however it can be useless when unneeded in smaller logarithms. In terms of a, b and x, x must repeat in order to use this expression. To find the general statement of cd/ c+d, was by reverse solution. I found the answer of the third term and tried to find out how this answer could be derived by using the values calculated from the first two terms. Word count: 1475 ?? ?? ?? ?? Elliott 1 ...read more.

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